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Classic examples of C language 100

2022-06-24 17:04:00 【C language and CPP programming】

source ： official account （c Language and cpp Programming ）, The background to reply “100” obtain pdf

【 Program 1】

subject ： Yes 1、2、3、4 A digital , How many different and unrepeated three digit numbers can be formed ？ How many are they ？

Program analysis ： It can be filled in the hundreds 、 ten 、 All the figures in a single digit are 1、2、3、4. Form all permutations and then remove the ones that don't meet the conditions .
Program source code ：

int _tmain(int argc, _TCHAR* argv[])
{
 int i, j, k,n=0;
 for (i = 1; i < 5; i++)
  for (j = 1; j < 5; j++)
   for (k = 1; k < 5; k++)
   {
    if (i != k&&i != j&&j != k){
     n++;
     printf("%d%d%d\n", i, j, k);
    }   
   }
 printf(" Altogether %d individual \n", n);
 system("pause");
 return 0;
}

【 Program 2】

subject ： The bonus paid by the enterprise is based on the profit commission .

profits (I) Below or equal to 10 Ten thousand yuan , Bonus can be raised 10%;
Profit is higher than 10 Ten thousand yuan , lower than 20 Ten thousand yuan , lower than 10 For the part of ten thousand yuan, press 10% Royalty , higher than 10 Part of ten thousand yuan , Cocoa Commission 7.5%;
20 Wan to 40 Ten thousand hours , higher than 20 Part of ten thousand yuan , It's a percentage 5%;
40 Wan to 60 Ten thousand times is higher than 40 Part of ten thousand yuan , It's a percentage 3%;
60 Wan to 100 Ten thousand hours , higher than 60 Part of ten thousand yuan , It's a percentage 1.5%,
higher than 100 Ten thousand yuan , exceed 100 For the part of ten thousand yuan, press 1% Royalty , Input the profit of the month from the keyboard I, Ask for the total amount of bonus payable ？

Program analysis ： Please use the number axis to divide , location . Pay attention to defining bonus as growth integer .
Program source code ：

main()
{
long int i;
int bonus1,bonus2,bonus4,bonus6,bonus10,bonus;
scanf("%ld",&i);
bonus1=100000*0.1;bonus2=bonus1+100000*0.75;
bonus4=bonus2+200000*0.5;
bonus6=bonus4+200000*0.3;
bonus10=bonus6+400000*0.15;
  if(i<=100000)
   bonus=i*0.1;
  else if(i<=200000)
      bonus=bonus1+(i-100000)*0.075;
     else if(i<=400000)
         bonus=bonus2+(i-200000)*0.05;
        else if(i<=600000)
            bonus=bonus4+(i-400000)*0.03;
           else if(i<=1000000)
               bonus=bonus6+(i-600000)*0.015;
              else
               bonus=bonus10+(i-1000000)*0.01;
printf("bonus=%d",bonus);
}

【 Program 3】

subject ： An integer , It adds 100 And then there's a complete square , Plus 168 It's a complete square again , What is the number ？

Program analysis ： stay 10 Judge within ten thousand , First add the number to 100 We'll start the recipe later , Add this number to 268 We'll start the recipe later , If the result of the prescription meets the following conditions , It's the result . Please see the specific analysis ：
Program source code ：

#include "math.h"
main()
{
long int i,x,y,z;
for (i=1;i<100000;i++)
  { x=sqrt(i+100);   /*x To add 100 The result of the later prescription is */
   y=sqrt(i+268);   /*y To add 168 The result of the later prescription is */
    if(x*x==i+100&&y*y==i+268)/* If the square of the square root of a number is equal to the number , This shows that the number is a complete square */
     printf("\n%ld\n",i);
  }
}

【 Program 4】

subject ： Enter the date of the year , Judge the day as the day of the year ？

Program analysis ： With 3 month 5 Day, for example , The first two months should be added up , And then add 5 Day is the day of the year , A special case , Leap year and input month is greater than 3 Consider an extra day .
Program source code ：

int _tmain(int argc, _TCHAR* argv[])
{
 int day, month, year, sum, leap;
 printf("please input year,month,day:\n");
 scanf_s("%d %d %d", &year, &month, &day);
 switch (month)/* First, calculate the total number of days in the month before a certain month */
 {
 case 1:sum = 0; break;
 case 2:sum = 31; break;
 case 3:sum = 59; break;
 case 4:sum = 90; break;
 case 5:sum = 120; break;
 case 6:sum = 151; break;
 case 7:sum = 181; break;
 case 8:sum = 212; break;
 case 9:sum = 243; break;
 case 10:sum = 273; break;
 case 11:sum = 304; break;
 case 12:sum = 334; break;
 default:printf("data error"); break;
 }
 sum += day;  /* Plus the days of the day */
 if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))/* Decide if it's a leap year */
 {
  leap = 1;
 }else{
  leap = 0;
 }
 if (leap == 1 && month > 2)/* If it is a leap year and the month is greater than 2, The total number of days should be increased by one day */
 {
  sum++;
 }
 printf("It is the %d th day.", sum);
 system("pause");
 return 0;
}

【 Program 5】

subject ： Enter three integers x,y,z, Please output these three numbers from small to large .

Program analysis ： We try to put the minimum number in x On , First the x And y Compare , If x>y Will x And y The value of , And then use x And z Compare , If x>z Will x And z The value of , This can make x Minimum .
Program source code ：

main()
{
int x,y,z,t;
scanf("%d%d%d",&x,&y,&z);
if (x>y)
{t=x;x=y;y=t;} /* In exchange for x,y Value */
if(x>z)
{t=z;z=x;x=t;}/* In exchange for x,z Value */
if(y>z)
{t=y;y=z;z=t;}/* In exchange for z,y Value */
printf("small to big: %d %d %d\n",x,y,z);
}

【 Program 6】

seek 100 The prime number within

Program analysis ：
Program source code ：

#include "stdio.h"
#include "math.h"
#define N 101
main()
{
  int i,j,line,a[N];
  for(i=2;i<N;i++) a[i]=i;
    for(i=2;i<sqrt(N);i++)
      for(j=i+1;j<N;j++)
      {
        if(a[i]!=0&&a[j]!=0)
          if(a[j]%a[i]==0)
            a[j]=0;
      }
  printf("\n");
  for(i=2,line=0;i<N;i++)
  {
    if(a[i]!=0)
    {
      printf("%5d",a[i]);
      line++;
    }
    if(line==10)
    {
      printf("\n");
      line=0;
    }
  }
  getch();
}

【 Program 7】

subject ： Yes 10 Number to sort

Program analysis ： You can use the method of choice , From the back 9 In the process of comparison , Choose the smallest to exchange with the first element , And so on next time , Use the second element and the post 8 Compare them , And exchange .
Program source code ：

#include "stdio.h"
#include "conio.h"
#define N 10
main()
{
  int i,j,min,tem,a[N];
  /*input data*/
  printf("please input ten num:\n");
  for(i=0;i<N;i++)
  {
    printf("a[%d]=",i);
    scanf("%d",&a[i]);
  }
  printf("\n");
  for(i=0;i<N;i++)
    printf("%5d",a[i]);
  printf("\n");
  /*sort ten num*/
  for(i=0;i<N-1;i++)
  {
    min=i;
    for(j=i+1;j<N;j++)
      if(a[min]>a[j])
        min=j;
    tem=a[i];
    a[i]=a[min];
    a[min]=tem;
  }
  /*output data*/
  printf("After sorted \n");
  for(i=0;i<N;i++)
  printf("%5d",a[i]);
  getch();
}

【 Program 8】

subject ： Output 9*9 Multiplication table

int main()
{
 int x, y, z;
 for (x = 0; x < 9; x++)
 {
  for (y = 0; y < x; y++)
  {
   z = x*y;
   printf("%d*%d=%d  ", y, x, z);
  }
  printf("\n");
 }
 system("pause");
 return 0;
}

【 Program 9】

subject ： Yes 5 A person sits together , Ask the fifth person how old ？ He said bidi 4 Personal big 2 year . Ask No 4 Personal age , He said bidi 3 Personal big 2 year . Ask the third person , And he said bidi 2 People are two years old . Ask No 2 personal , Say two years older than the first one . Finally, ask the first person , He says it is 10 year . How old is the fifth person ？

Program analysis ： Using recursive methods , Recursion can be divided into two stages: backward and recursive . Want to know the age of the fifth person , You need to know the age of the fourth person , By analogy , Push to the first person （10 year ）, Push back .
Program source code ：

#include "stdio.h"
#include "conio.h"
age(n)
int n;
{
  int c;
  if(n==1) c=10;
  else c=age(n-1)+2;
  return(c);
}
main()
{
  printf("%d",age(5));
  getch();
}

【 Program 10】

subject ： There's a sequence of scores ：2/1,3/2,5/3,8/5,13/8,21/13... Find the front of this sequence 20 Sum of items .

Program analysis ： Please grasp the law of change of the numerator and denominator .
Program source code ：

#include "stdio.h"
#include "conio.h"
main()
{
  int n,t,number=20;
  float a=2,b=1,s=0;
  for(n=1;n<=number;n++)
  {
    s=s+a/b;
    t=a;a=a+b;b=t;/* This part is the key to the program , Please guess t The role of */
  }
  printf("sum is %9.6f\n",s);
  getch();
}

【 Program 11】

subject ： Classical questions ： There is a pair of rabbits , From the day after birth 3 A couple of rabbits are born every month from , Every month after the third month, a couple of rabbits will be born , If the rabbits don't die , Ask the total number of rabbits per month ？

Program analysis ：　 The law of the rabbit is a series 1,1,2,3,5,8,13,21....
Program source code ：

main()
{
long f1,f2;
int i;
f1=f2=1;
for(i=1;i<=20;i++)
  { printf("%12ld %12ld",f1,f2);
    if(i%2==0) printf("\n");/* Control output , Four in each line */
    f1=f1+f2; /* Add up the first two months and assign the value to the third month */
    f2=f1+f2; /* Add up the first two months and assign the value to the third month */
  }
}

【 Program 12】

subject ： Judge 101-200 How many primes are there between , And output all prime numbers .

Program analysis ： The way to judge prime numbers ： Remove with a number 2 To sqrt( The number of ), If it can be divided , It means that this number is not a prime number , On the contrary, prime numbers .
Program source code ：

#include "math.h"
main()
{
  int m,i,k,h=0,leap=1;
  printf("\n");
  for(m=101;m<=200;m++)
   { k=sqrt(m+1);
    for(i=2;i<=k;i++)
      if(m%i==0)
       {leap=0;break;}
    if(leap) {printf("%-4d",m);h++;
         if(h%10==0)
         printf("\n");
        }
    leap=1;
   }
  printf("\nThe total is %d",h);
}

【 Program 13】

subject ： Print out all “ Narcissistic number ”, So-called “ Narcissistic number ” A three digit number , The sum of its cubes is equal to the number itself . for example ：153 It's a “ Narcissistic number ”, because 153=1 The third power of ＋5 The third power of ＋3 The third power of .

Program analysis ： utilize for Cycle control 100-999 Number , Each number is decomposed into bits , ten , Hundred bit .
Program source code ：

main()
{
int i,j,k,n;
printf("'water flower'number is:");
  for(n=100;n<1000;n++)
  {
   i=n/100;/* Break down into hundreds */
   j=n/10%10;/* Break it down into ten */
   k=n%10;/* Break down a bit */
   if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)
    {
    printf("%-5d",n);
    }
  }
printf("\n");
}

【 Program 14】

subject ： Divide a positive integer into prime factors . for example ： Input 90, Print out 90=233*5.

Program analysis ： Yes n To decompose the prime factor , We should find a minimum prime number first k, Then follow the steps below ：

If this prime is exactly equal to n, Then the process of decomposing the prime factor is over , Just print it out .
If n<>k, but n Can be k to be divisible by , Then print out k Value , And use n Divide k The business of , As a new positive integer you n, 　 Repeat the first step .
If n Can not be k to be divisible by , Then use k+1 As k Value , Repeat the first step .

Program source code ：

/* zheng int is divided yinshu*/
main()
{
int n,i;
printf("\nplease input a number:\n");
scanf("%d",&n);
printf("%d=",n);
for(i=2;i<=n;i++)
  {
   while(n!=i)
   {
    if(n%i==0)
    { printf("%d*",i);
     n=n/i;
    }
    else
     break;
   }
}
printf("%d",n);}

【 Program 15】

subject ： Using the nesting of conditional operators to complete this problem ： academic record >=90 It's for the students A Express ,60-89 Use between points B Express ,60 Divide the following uses C Express .

Program analysis ：(a>b)?a:b This is a basic example of a conditional operator .
Program source code ：

main()
{
  int score;
  char grade;
  printf("please input a score\n");
  scanf("%d",&score);
  grade=score>=90?'A':(score>=60?'B':'C');
  printf("%d belongs to %c",score,grade);
}

【 Program 16】

subject ： Enter two positive integers m and n, Find the greatest common divisor and the least common multiple .

Program analysis ： Use the method of rolling .
Program source code ：

main()
{
  int a,b,num1,num2,temp;
  printf("please input two numbers:\n");
  scanf("%d,%d",&num1,&num2);
  if(num1<num2)   
 { temp=num1;
   num1=num2; 
   num2=temp;
  }
a=num1;b=num2;
while(b!=0)/* Use the method of rolling , until b by 0 until */
  {
   temp=a%b;
   a=b;
   b=temp;
  }
printf("gongyueshu:%d\n",a);
printf("gongbeishu:%d\n",num1*num2/a);
}

【 Program 17】

subject ： Enter a line of characters , Count out the English letters 、 Space 、 The number of numbers and other characters .

Program analysis ： utilize while sentence , The condition is that the character entered is not '\n'.
Program source code ：

#include "stdio.h"
main()
{char c;
  int letters=0,space=0,digit=0,others=0;
  printf("please input some characters\n");
  while((c=getchar())!='\n')
  {
  if(c>='a'&&c<='z'||c>='A'&&c<='Z')
   letters++;
  else if(c==' ')
   space++;
    else if(c>='0'&&c<='9')
        digit++;
      else
        others++;
}
printf("all in all:char=%d space=%d digit=%d others=%d\n",letters,
space,digit,others);
}

【 Program 18】

subject ： seek s=a+aa+aaa+aaaa+aa...a Value , among a It's a number . for example 2+22+222+2222+22222( At this time, there is 5 Add up the numbers ), A keyboard controls the addition of several numbers .

Program analysis ： The key is to calculate the value of each item .
Program source code ：

main()
{
  int a,n,count=1;
  long int sn=0,tn=0;
  printf("please input a and n\n");
  scanf("%d,%d",&a,&n);
  printf("a=%d,n=%d\n",a,n);
  while(count<=n)
  {
   tn=tn+a;
   sn=sn+tn;
   a=a*10;
   ++count;
  }
printf("a+aa+...=%ld\n",sn);
}

【 Program 19】

subject ： If a number is exactly equal to the sum of its factors , This number is called “ Complete ”. for example 6=1＋2＋3. Programming to find out 1000 All the completions within .

Program analysis ： Please refer to the procedure <-- The program on the previous page 14.
Program source code ：

main()
{
static int k[10];
int i,j,n,s;
for(j=2;j<1000;j++)
  {
  n=-1;
  s=j;
   for(i=1;i<j;i++)
  {
    if((j%i)==0)
    { n++;
     s=s-i;
     k[n]=i;
    }
   }
  if(s==0)
  {
  printf("%d is a wanshu",j);
  for(i=0;i<n;i++)   
 printf("%d,",k[i]);
  printf("%d\n",k[n]);
  }
}
}

【 Program 20】

subject ： A ball from 100 Free fall at meter height , Jump back to half of the original height after landing ; And then fall , Ask it in the 10 The next landing , How many meters in total ？ The first 10 How high is the rebound ？

Program analysis ： See note below
Program source code ：

main()
{
float sn=100.0,hn=sn/2;
int n;
for(n=2;n<=10;n++)
  {
   sn=sn+2*hn;/* The first n The total number of meters passed during the first landing */
   hn=hn/2; /* The first n Second bounce height */
  }
printf("the total of road is %f\n",sn);
printf("the tenth is %f meter\n",hn);
}

【 Program 21】

subject ： The problem of monkeys eating peaches ： The monkey picked some peaches on the first day , Half eaten immediately , It's not addictive , Another one , The next morning I ate half of the rest of the peaches , Another one . After that, I ate half and one of the rest of the day before every morning . To the first 10 When I want to eat again in the morning , See there's only one peach left . Ask how much you picked on the first day .

Program analysis ： Adopt the method of converse thinking , Infer from back to front .
Program source code ：

main()
{
int day,x1,x2;
day=9;
x2=1;
while(day>0)
  {x1=(x2+1)*2;/* The number of peaches on the first day was 2 The number of peaches in the sky 1 After 2 times */
  x2=x1;
  day--;
  }
printf("the total is %d\n",x1);
}

【 Program 22】

subject ： Two ping-pong teams play , Three people each . Team a is for a,b,c Three people , Team B is x,y,z Three people . The match list has been drawn . Someone asked the players for the list of the game .a Say he's not at peace x Than ,c Say he's not at peace x,z Than , Please program to find out the players of the three teams .

Program analysis ： The way to judge prime numbers ： Remove with a number 2 To sqrt( The number of ), If it can be divided , It means that this number is not a prime number , On the contrary, prime numbers .
Program source code ：

main()
{
char i,j,k;/*i yes a The opponent ,j yes b The opponent ,k yes c The opponent */
for(i='x';i<='z';i++)
  for(j='x';j<='z';j++)
  {
  if(i!=j)
   for(k='x';k<='z';k++)
   { if(i!=k&&j!=k)
    { if(i!='x'&&k!='x'&&k!='z')
    printf("order is a--%c\tb--%c\tc--%c\n",i,j,k);
    }
   }
  }
}

【 Program 23】

subject ： Print out the following pattern （ The diamond ） *

Program analysis ： First, divide the figure into two parts , One rule in the first four lines , The last three lines have one rule , Use double for loop , First level control line , The second control column .
Program source code ：

main()
{
int i,j,k;
for(i=0;i<=3;i++)
  {
  for(j=0;j<=2-i;j++)
   printf(" ");
  for(k=0;k<=2*i;k++)
   printf("*");
  printf("\n");
  }
for(i=0;i<=2;i++)
  {
  for(j=0;j<=i;j++)
   printf(" ");
  for(k=0;k<=4-2*i;k++)
   printf("*");
  printf("\n");
  }
}

【 Program 24】

subject ： There's a sequence of scores ：2/1,3/2,5/3,8/5,13/8,21/13... Find the front of this sequence 20 Sum of items .

Program analysis ： Please grasp the law of change of the numerator and denominator .
Program source code ：

main()
{
int n,t,number=20;
float a=2,b=1,s=0;
for(n=1;n<=number;n++)
  {
  s=s+a/b;
  t=a;a=a+b;b=t;/* This part is the key to the program , Please guess t The role of */
  }
printf("sum is %9.6f\n",s);
}

【 Program 25】

subject ： seek 1+2!+3!+...+20! And

Program analysis ： This program just turns accumulation into multiplication .
Program source code ：

main()
{
float n,s=0,t=1;
for(n=1;n<=20;n++)
  {
  t*=n;
  s+=t;
  }
printf("1+2!+3!...+20!=%e\n",s);
}

【 Program 26】

subject ： Use recursion to find 5!.

Program analysis ： Recursive formula ：fn=fn_1*4!
Program source code ：

#include "stdio.h"
main()
{
int i;
int fact();
for(i=0;i<5;i++)
  printf("\40:%d!=%d\n",i,fact(i));
}
int fact(j)
int j;
{
int sum;
if(j==0)
  sum=1;
else
  sum=j*fact(j-1);
return sum;
}

【 Program 27】

subject ： Using recursive function call mode , Put in 5 Characters , Print out in reverse order .

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{
int i=5;
void palin(int n);
printf("\40:");
palin(i);
printf("\n");
}
void palin(n)
int n;
{
char next;
if(n<=1)
  {
  next=getchar();
  printf("\n\0:");
  putchar(next);
  }
else
  {
  next=getchar();
  palin(n-1);
  putchar(next);
  }
}

【 Program 28】

Program analysis ： Using recursive methods , Recursion can be divided into two stages: backward and recursive . Want to know the age of the fifth person , You need to know the age of the fourth person , By analogy , Push to the first person （10 year ）, Push back .
Program source code ：

age(n)
int n;
{
int c;
if(n==1) c=10;
else c=age(n-1)+2;
return(c);
}
main()
{ printf("%d",age(5));
}

【 Program 29】

subject ： Give one not more than 5 Bit positive integer , requirement ： One 、 Find out how many digits it is , Two 、 Print out the numbers in reverse order .

Program analysis ： Learn to break down every digit
Program source code ：

main( )
{
long a,b,c,d,e,x;
scanf("%ld",&x);
a=x/10000;/* Decompose ten thousand */
b=x%10000/1000;/* Decompose into thousands */
c=x%1000/100;/* Break down into hundreds */
d=x%100/10;/* Break it down into ten */
e=x%10;/* Break down a bit */
if (a!=0) printf("there are 5, %ld %ld %ld %ld %ld\n",e,d,c,b,a);
else if (b!=0) printf("there are 4, %ld %ld %ld %ld\n",e,d,c,b);
   else if (c!=0) printf(" there are 3,%ld %ld %ld\n",e,d,c);
     else if (d!=0) printf("there are 2, %ld %ld\n",e,d);
       else if (e!=0) printf(" there are 1,%ld\n",e);
}

【 Program 30】

subject ： One 5 digit , Judge whether it is palindrome number . namely 12321 It's palindrome number , One digit is the same as ten thousand , Ten is the same as one thousand .

Program analysis ： Same as 29 example
Program source code ：

main( )
{
long ge,shi,qian,wan,x;
scanf("%ld",&x);
wan=x/10000;
qian=x%10000/1000;
shi=x%100/10;
ge=x%10;
if (ge==wan&&shi==qian)/* One digit equals ten thousand and ten digit equals one thousand */
  printf("this number is a huiwen\n");
else
  printf("this number is not a huiwen\n");
}

【 Program 31】

subject ： Please input the first letter of the day of the week to determine the day of the week , If the first letter is the same , Then continue to judge the second letter .

Program analysis ： It's better to use situation statements , If the first letter is the same , Then judge the situation statement or if Sentence judge the second letter .
Program source code ：

#include <stdio.h>
void main()
{
char letter;
printf("please input the first letter of someday\n");
while ((letter=getch())!='Y')/* When the letter is Y It doesn't end until */
{ switch (letter)
{case 'S':printf("please input second letter\n");
      if((letter=getch())=='a')
       printf("saturday\n");
      else if ((letter=getch())=='u')
          printf("sunday\n");
        else printf("data error\n");
      break;
case 'F':printf("friday\n");break;
case 'M':printf("monday\n");break;
case 'T':printf("please input second letter\n");
      if((letter=getch())=='u')
       printf("tuesday\n");
      else if ((letter=getch())=='h')
          printf("thursday\n");
        else printf("data error\n");
      break;
case 'W':printf("wednesday\n");break;
default: printf("data error\n");
   }
  }
}

【 Program 32】

subject ：Press any key to change color, do you want to try it. Please hurry up!

Program analysis ：
Program source code ：

#include <conio.h>
void main(void)
{
int color;
for (color = 0; color < 8; color++)
  { 
  textbackground(color);/* Set the background color of the text */
  cprintf("This is color %d\r\n", color);
  cprintf("Press any key to continue\r\n");
  getch();/* Input characters cannot be seen */
  }
}

【 Program 33】

subject ： Study gotoxy() And clrscr() function

Program analysis ：
Program source code ：

#include <conio.h>
void main(void)
{
clrscr();/* Screen clearing function */
textbackground(2);
gotoxy(1, 5);/* Location function */
cprintf("Output at row 5 column 1\n");
textbackground(3);
gotoxy(20, 10);
cprintf("Output at row 10 column 20\n");
}

【 Program 34】

subject ： Practice function calls

Program analysis ：
Program source code ：

#include <stdio.h>
void hello_world(void)
{
printf("Hello, world!\n");
}
void three_hellos(void)
{
int counter;
for (counter = 1; counter <= 3; counter++)
hello_world();/* Call this function */
}
void main(void)
{
three_hellos();/* Call this function */
}

【 Program 35】

subject ： Text color settings

Program analysis ：
Program source code ：

#include <conio.h>
void main(void)
{
int color;
for (color = 1; color < 16; color++)
  {
  textcolor(color);/* Set text color */
  cprintf("This is color %d\r\n", color);
  }
textcolor(128 + 15);
cprintf("This is blinking\r\n");
}

【 Program 36】

subject ： seek 100 The prime number within

Program analysis ：
Program source code ：

#include <stdio.h>
#include "math.h"
#define N 101
main()
{
int i,j,line,a[N];
for(i=2;i<N;i++) a[i]=i;
for(i=2;i<sqrt(N);i++)
  for(j=i+1;j<N;j++)
  {
   if(a[i]!=0&&a[j]!=0)
   if(a[j]%a[i]==0)
   a[j]=0;}
printf("\n");
for(i=2,line=0;i<N;i++)
{
  if(a[i]!=0)
  {printf("%5d",a[i]);
  line++;}
  if(line==10)
  {printf("\n");
line=0;}
}
}

【 Program 37】

subject ： Yes 10 Number to sort

Program analysis ： You can use the method of choice , From the back 9 In the process of comparison , Choose the smallest to exchange with the first element , And so on next time , Use the second element and the post 8 Compare them , And exchange .
Program source code ：

#define N 10
main()
{int i,j,min,tem,a[N];
/*input data*/
printf("please input ten num:\n");
for(i=0;i<N;i++)
{
printf("a[%d]=",i);
scanf("%d",&a[i]);}
printf("\n");
for(i=0;i<N;i++)
printf("%5d",a[i]);
printf("\n");
/*sort ten num*/
for(i=0;i<N-1;i++)
{min=i;
for(j=i+1;j<N;j++)
if(a[min]>a[j]) min=j;
tem=a[i];
a[i]=a[min];
a[min]=tem;
}
/*output data*/
printf("After sorted \n");
for(i=0;i<N;i++)
printf("%5d",a[i]);
}

【 Program 38】

subject ： Ask for one 3*3 The sum of diagonal elements of a matrix

Program analysis ： Use double for Loop control input two-dimensional array , then a[i][i] Output after accumulation .
Program source code ：

main()
{
float a[3][3],sum=0;
int i,j;
printf("please input rectangle element:\n");
for(i=0;i<3;i++)
  for(j=0;j<3;j++)
  scanf("%f",&a[i][j]);
for(i=0;i<3;i++)
  sum=sum+a[i][i];
printf("duijiaoxian he is %6.2f",sum);
}

【 Program 39】

subject ： There's an array that's already in order . Now enter a number , It is required to insert it into the array according to the original rule .

Program analysis ： First, judge whether the number is greater than the last number , Then consider inserting the middle number , The number after this element is inserted , Move back one position in turn .
Program source code ：

main()
{
int a[11]={1,4,6,9,13,16,19,28,40,100};
int temp1,temp2,number,end,i,j;
printf("original array is:\n");
for(i=0;i<10;i++)
  printf("%5d",a[i]);
printf("\n");
printf("insert a new number:");
scanf("%d",&number);
end=a[9];
if(number>end)
  a[10]=number;
else
  {for(i=0;i<10;i++)
   { if(a[i]>number)
    {temp1=a[i];
     a[i]=number;
    for(j=i+1;j<11;j++)
    {temp2=a[j];
     a[j]=temp1;
     temp1=temp2;
    }
    break;
    }
   }
}
for(i=0;i<11;i++)
  printf("%6d",a[i]);
}

【 Program 40】

subject ： Output an array in reverse order .

Program analysis ： Exchange the first with the last .
Program source code ：

#define N 5
main()
{ int a[N]={9,6,5,4,1},i,temp;
  printf("\n original array:\n");
  for(i=0;i<N;i++)
  printf("%4d",a[i]);
  for(i=0;i<N/2;i++)
  {temp=a[i];
   a[i]=a[N-i-1];
   a[N-i-1]=temp;
  }
printf("\n sorted array:\n");
for(i=0;i<N;i++)
  printf("%4d",a[i]);
}

【 Program 41】

subject ： Study static Define the usage of static variables

Program analysis ：
Program source code ：

#include "stdio.h"
varfunc()
{
int var=0;
static int static_var=0;
printf("\40:var equal %d \n",var);
printf("\40:static var equal %d \n",static_var);
printf("\n");
var++;
static_var++;
}
void main()
{int i;
  for(i=0;i<3;i++)
   varfunc();
}

【 Program 42】

subject ： Learn to use auto Define the use of variables

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{int i,num;
num=2;
  for (i=0;i<3;i++)
  { printf("\40: The num equal %d \n",num);
   num++;
   {
   auto int num=1;
   printf("\40: The internal block num equal %d \n",num);
   num++;
   }
  }
}

【 Program 43】

subject ： Learn to use static Another use of .

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{
int i,num;
num=2;
for(i=0;i<3;i++)
{
printf("\40: The num equal %d \n",num);
num++;
{
static int num=1;
printf("\40:The internal block num equal %d\n",num);
num++;
}
}
}

【 Program 44】

subject ： Learn to use external Usage of .

Program analysis ：
Program source code ：

#include "stdio.h"
int a,b,c;
void add()
{ int a;
a=3;
c=a+b;
}
void main()
{ a=b=4;
add();
printf("The value of c is equal to %d\n",c);
}

【 Program 45】

subject ： Learn to use register How to define variables .

Program analysis ：
Program source code ：

void main()
{
register int i;
int tmp=0;
for(i=1;i<=100;i++)
tmp+=i;
printf("The sum is %d\n",tmp);
}

【 Program 46】

subject ： macro #define Command practice (1)

Program analysis ：
Program source code ：

#include "stdio.h"
#define TRUE 1
#define FALSE 0
#define SQ(x) (x)*(x)
void main()
{
int num;
int again=1;
printf("\40: Program will stop if input value less than 50.\n");
while(again)
{
printf("\40:Please input number==>");
scanf("%d",&num);
printf("\40:The square for this number is %d \n",SQ(num));
if(num>=50)
  again=TRUE;
else
  again=FALSE;
}
}

【 Program 47】

subject ： macro #define Command practice (2)

Program analysis ：
Program source code ：

#include "stdio.h"
#define exchange(a,b) { \ /* It is allowed to include two clothes commands in the macro definition , At this point, you must add... To the far right "\"*/
             int t;\
             t=a;\
             a=b;\
             b=t;\
            }
void main(void)
{
int x=10;
int y=20;
printf("x=%d; y=%d\n",x,y);
exchange(x,y);
printf("x=%d; y=%d\n",x,y);
}

【 Program 48】

subject ： macro #define Command practice (3)

Program analysis ：
Program source code ：

#define LAG >
#define SMA <
#define EQ ==
#include "stdio.h"
void main()
{ int i=10;
int j=20;
if(i LAG j)
printf("\40: %d larger than %d \n",i,j);
else if(i EQ j)
printf("\40: %d equal to %d \n",i,j);
else if(i SMA j)
printf("\40:%d smaller than %d \n",i,j);
else
printf("\40: No such value.\n");
}

【 Program 49】

subject ：#if #ifdef and #ifndef Comprehensive application of .

Program analysis ：
Program source code ：

#include "stdio.h"
#define MAX
#define MAXIMUM(x,y) (x>y)?x:y
#define MINIMUM(x,y) (x>y)?y:x
void main()
{ int a=10,b=20;
#ifdef MAX
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
#else
printf("\40: The lower one is %d\n",MINIMUM(a,b));
#endif
#ifndef MIN
printf("\40: The lower one is %d\n",MINIMUM(a,b));
#else
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
#endif
#undef MAX
#ifdef MAX
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
#else
printf("\40: The lower one is %d\n",MINIMUM(a,b));
#endif
#define MIN
#ifndef MIN
printf("\40: The lower one is %d\n",MINIMUM(a,b));
#else
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
#endif
}

【 Program 50】

subject ：#include Practical exercises of

Program analysis ：
Program source code ： test.h The documents are as follows ：

#define LAG >
#define SMA <
#define EQ ==
#include "test.h" /* A new document 50.c, contain test.h*/
#include "stdio.h"
void main()
{ int i=10;
int j=20;
if(i LAG j)
printf("\40: %d larger than %d \n",i,j);
else if(i EQ j)
printf("\40: %d equal to %d \n",i,j);
else if(i SMA j)
printf("\40:%d smaller than %d \n",i,j);
else
printf("\40: No such value.\n");
}

【 Program 51】

subject ： Learn to use place and & .

Program analysis ：0&0=0; 0&1=0; 1&0=0; 1&1=1
Program source code ：

#include "stdio.h"
main()
{
int a,b;
a=077;
b=a&3;
printf("\40: The a & b(decimal) is %d \n",b);
b&=7;
printf("\40: The a & b(decimal) is %d \n",b);
}

【 Program 52】

subject ： Learn to use bitwise OR | .

Program analysis ：0|0=0; 0|1=1; 1|0=1; 1|1=1
Program source code ：

#include "stdio.h"
main()
{
int a,b;
a=077;
b=a|3;
printf("\40: The a & b(decimal) is %d \n",b);
b|=7;
printf("\40: The a & b(decimal) is %d \n",b);
}

【 Program 53】

subject ： Learn to use bitwise XOR ^ .

Program analysis ：0^0=0; 0^1=1; 1^0=1; 1^1=0
Program source code ：

#include "stdio.h"
main()
{
int a,b;
a=077;
b=a^3;
printf("\40: The a & b(decimal) is %d \n",b);
b^=7;
printf("\40: The a & b(decimal) is %d \n",b);
}

【 Program 54】

subject ： Take an integer a From the right end 4～7 position .

Program analysis ： Think about it like this ：

First of all a Move right 4 position .
Set a low 4 All the seats are 1, The rest are all 0 Number of numbers . You can use ~(~0<<4)
Do both of the above & operation .

Program source code ：

main()
{
unsigned a,b,c,d;
scanf("%o",&a);
b=a>>4;
c=~(~0<<4);
d=b&c;
printf("%o\n%o\n",a,d);
}

【 Program 55】

subject ： Learn to use bitwise negation ~.

Program analysis ：~0=1; ~1=0;
Program source code ：

#include "stdio.h"
main()
{
int a,b;
a=234;
b=~a;
printf("\40: The a's 1 complement(decimal) is %d \n",b);
a=~a;
printf("\40: The a's 1 complement(hexidecimal) is %x \n",a);
}

【 Program 56】

subject ： drawing , Learn to use circle Draw a circle .

Program analysis ：
Program source code ：

/*circle*/
#include "graphics.h"
main()
{int driver,mode,i;
float j=1,k=1;
driver=VGA;mode=VGAHI;
initgraph(&driver,&mode,"");
setbkcolor(YELLOW);
for(i=0;i<=25;i++)
{
setcolor(8);
circle(310,250,k);
k=k+j;
j=j+0.3;
}
}

【 Program 57】

subject ： drawing , Learn to use line Draw a straight line .

Program analysis ：
Program source code ：

#include "graphics.h"
main()
{int driver,mode,i;
float x0,y0,y1,x1;
float j=12,k;
driver=VGA;mode=VGAHI;
initgraph(&driver,&mode,"");
setbkcolor(GREEN);
x0=263;y0=263;y1=275;x1=275;
for(i=0;i<=18;i++)
{
setcolor(5);
line(x0,y0,x0,y1);
x0=x0-5;
y0=y0-5;
x1=x1+5;
y1=y1+5;
j=j+10;
}
x0=263;y1=275;y0=263;
for(i=0;i<=20;i++)
{
setcolor(5);
line(x0,y0,x0,y1);
x0=x0+5;
y0=y0+5;
y1=y1-5;
}
}

【 Program 58】

subject ： drawing , Learn to use rectangle Draw a square .

Program analysis ： utilize for Cycle control 100-999 Number , Each number is decomposed into bits , ten , Hundred bit .
Program source code ：

#include "graphics.h"
main()
{int x0,y0,y1,x1,driver,mode,i;
driver=VGA;mode=VGAHI;
initgraph(&driver,&mode,"");
setbkcolor(YELLOW);
x0=263;y0=263;y1=275;x1=275;
for(i=0;i<=18;i++)
{
setcolor(1);
rectangle(x0,y0,x1,y1);
x0=x0-5;
y0=y0-5;
x1=x1+5;
y1=y1+5;
}
settextstyle(DEFAULT_FONT,HORIZ_DIR,2);
outtextxy(150,40,"How beautiful it is!");
line(130,60,480,60);
setcolor(2);
circle(269,269,137);
}

【 Program 59】

subject ： drawing , A comprehensive example .

Program analysis ：
Program source code ：

# define PAI 3.1415926
# define B 0.809
# include "graphics.h"
#include "math.h"
main()
{
int i,j,k,x0,y0,x,y,driver,mode;
float a;
driver=CGA;mode=CGAC0;
initgraph(&driver,&mode,"");
setcolor(3);
setbkcolor(GREEN);
x0=150;y0=100;
circle(x0,y0,10);
circle(x0,y0,20);
circle(x0,y0,50);
for(i=0;i<16;i++)
{
  a=(2*PAI/16)*i;
  x=ceil(x0+48*cos(a));
  y=ceil(y0+48*sin(a)*B);
  setcolor(2); line(x0,y0,x,y);}
setcolor(3);circle(x0,y0,60);
/* Make 0 time normal size letters */
settextstyle(DEFAULT_FONT,HORIZ_DIR,0);
outtextxy(10,170,"press a key");
getch();
setfillstyle(HATCH_FILL,YELLOW);
floodfill(202,100,WHITE);
getch();
for(k=0;k<=500;k++)
{
  setcolor(3);
  for(i=0;i<=16;i++)
  {
   a=(2*PAI/16)*i+(2*PAI/180)*k;
   x=ceil(x0+48*cos(a));
   y=ceil(y0+48+sin(a)*B);
   setcolor(2); line(x0,y0,x,y);
  }
  for(j=1;j<=50;j++)
  {
   a=(2*PAI/16)*i+(2*PAI/180)*k-1;
   x=ceil(x0+48*cos(a));
   y=ceil(y0+48*sin(a)*B);
   line(x0,y0,x,y);
  }
}
restorecrtmode();
}

【 Program 60】 subject ： drawing , A comprehensive example .

Program analysis ：
Program source code ：

#include "graphics.h"
#define LEFT 0
#define TOP 0
#define RIGHT 639
#define BOTTOM 479
#define LINES 400
#define MAXCOLOR 15
main()
{
int driver,mode,error;
int x1,y1;
int x2,y2;
int dx1,dy1,dx2,dy2,i=1;
int count=0;
int color=0;
driver=VGA;
mode=VGAHI;
initgraph(&driver,&mode,"");
x1=x2=y1=y2=10;
dx1=dy1=2;
dx2=dy2=3;
while(!kbhit())
{
  line(x1,y1,x2,y2);
  x1+=dx1;y1+=dy1;
  x2+=dx2;y2+dy2;
  if(x1<=LEFT||x1>=RIGHT)
  dx1=-dx1;
  if(y1<=TOP||y1>=BOTTOM)
   dy1=-dy1;
  if(x2<=LEFT||x2>=RIGHT)
   dx2=-dx2;
  if(y2<=TOP||y2>=BOTTOM)
   dy2=-dy2;
  if(++count>LINES)
  {
   setcolor(color);
   color=(color>=MAXCOLOR)?0:++color;
  }
}
closegraph();
}

【 Program 61】

subject ： Print out Yang Hui triangle （ Ask to print out 10 The line is as follows ）

Program analysis ：　　　　　 1 　　　　　　 1 　1 　　　　　　 1 　2 　1 　　　　　　 1　 3 　3　 1 　　　　　　 1　 4　 6 　4 　1 　　　　　　 1　 5　 10　10　5 　1
Program source code ：

main()
{int i,j;
int a[10][10];
printf("\n");
for(i=0;i<10;i++)
  {a[i][0]=1;
  a[i][i]=1;}
for(i=2;i<10;i++)
  for(j=1;j<i;j++)
  a[i][j]=a[i-1][j-1]+a[i-1][j];
for(i=0;i<10;i++)
  {for(j=0;j<=i;j++)
  printf("%5d",a[i][j]);
  printf("\n");
  }
}

【 Program 62】

subject ： Study putpixel Draw a dot .

Program analysis ：
Program source code ：

#include "stdio.h"
#include "graphics.h"
main()
{
int i,j,driver=VGA,mode=VGAHI;
initgraph(&driver,&mode,"");
setbkcolor(YELLOW);
for(i=50;i<=230;i+=20)
  for(j=50;j<=230;j++)
  putpixel(i,j,1);
for(j=50;j<=230;j+=20)
  for(i=50;i<=230;i++)
  putpixel(i,j,1);
}

【 Program 63】

subject ： Drawing ellipse ellipse

Program analysis ：
Program source code ：

#include "stdio.h"
#include "graphics.h"
#include "conio.h"
main()
{
int x=360,y=160,driver=VGA,mode=VGAHI;
int num=20,i;
int top,bottom;
initgraph(&driver,&mode,"");
top=y-30;
bottom=y-30;
for(i=0;i<num;i++)
{
ellipse(250,250,0,360,top,bottom);
top-=5;
bottom+=5;
}
getch();
}

【 Program 64】

subject ： utilize ellipse and rectangle drawing .

Program analysis ：
Program source code ：

#include "stdio.h"
#include "graphics.h"
#include "conio.h"
main()
{
int driver=VGA,mode=VGAHI;
int i,num=15,top=50;
int left=20,right=50;
initgraph(&driver,&mode,"");
for(i=0;i<num;i++)
{
ellipse(250,250,0,360,right,left);
ellipse(250,250,0,360,20,top);
rectangle(20-2*i,20-2*i,10*(i+2),10*(i+2));
right+=5;
left+=5;
top+=10;
}
getch();
}

33 【 Program 65】 subject ： One of the most beautiful patterns .

Program analysis ：
Program source code ：

#include "graphics.h"
#include "math.h"
#include "dos.h"
#include "conio.h"
#include "stdlib.h"
#include "stdio.h"
#include "stdarg.h"
#define MAXPTS 15
#define PI 3.1415926
struct PTS {
int x,y;
};
double AspectRatio=0.85;
void LineToDemo(void)
{
struct viewporttype vp;
struct PTS points[MAXPTS];
int i, j, h, w, xcenter, ycenter;
int radius, angle, step;
double rads;
printf(" MoveTo / LineTo Demonstration" );
getviewsettings( &vp );
h = vp.bottom - vp.top;
w = vp.right - vp.left;
xcenter = w / 2; /* Determine the center of circle */
ycenter = h / 2;
radius = (h - 30) / (AspectRatio * 2);
step = 360 / MAXPTS; /* Determine # of increments */
angle = 0; /* Begin at zero degrees */
for( i=0 ; i<MAXPTS ; ++i ){ /* Determine circle intercepts */
rads = (double)angle * PI / 180.0; /* Convert angle to radians */
points[i].x = xcenter + (int)( cos(rads) * radius );
points[i].y = ycenter - (int)( sin(rads) * radius * AspectRatio );
angle += step; /* Move to next increment */
}
circle( xcenter, ycenter, radius ); /* Draw bounding circle */
for( i=0 ; i<MAXPTS ; ++i ){ /* Draw the cords to the circle */
for( j=i ; j<MAXPTS ; ++j ){ /* For each remaining intersect */
moveto(points[i].x, points[i].y); /* Move to beginning of cord */
lineto(points[j].x, points[j].y); /* Draw the cord */
} } }
main()
{int driver,mode;
driver=CGA;mode=CGAC0;
initgraph(&driver,&mode,"");
setcolor(3);
setbkcolor(GREEN);
LineToDemo();}

【 Program 66】

subject ： Input 3 Number a,b,c, Output... In order of size .

Program analysis ： Using the pointer method .
Program source code ：

/*pointer*/
main()
{
int n1,n2,n3;
int *pointer1,*pointer2,*pointer3;
printf("please input 3 number:n1,n2,n3:");
scanf("%d,%d,%d",&n1,&n2,&n3);
pointer1=&n1;
pointer2=&n2;
pointer3=&n3;
if(n1>n2) swap(pointer1,pointer2);
if(n1>n3) swap(pointer1,pointer3);
if(n2>n3) swap(pointer2,pointer3);
printf("the sorted numbers are:%d,%d,%d\n",n1,n2,n3);
}
swap(p1,p2)
int *p1,*p2;
{int p;
p=*p1;*p1=*p2;*p2=p;
}

【 Program 67】

subject ： Input array , The biggest exchange with the first element , The smallest exchange with the last element , The output array .

Program analysis ： There are questions in Tan Haoqiang's book .
Program source code ：

main()
{
int number[10];
input(number);
max_min(number);
output(number);
}
input(number)
int number[10];
{int i;
for(i=0;i<9;i++)
  scanf("%d,",&number[i]);
  scanf("%d",&number[9]);
}
max_min(array)
int array[10];
{int *max,*min,k,l;
int *p,*arr_end;
arr_end=array+10;
max=min=array;
for(p=array+1;p<arr_end;p++)
  if(*p>*max) max=p;
  else if(*p<*min) min=p;
  k=*max;
  l=*min;
  *p=array[0];array[0]=l;l=*p;
  *p=array[9];array[9]=k;k=*p;
  return;
}
output(array)
int array[10];
{ int *p;
for(p=array;p<array+9;p++)
  printf("%d,",*p);
printf("%d\n",array[9]);
}

【 Program 68】

subject ： Yes n It's an integer , Move the numbers back in order m A place , Last m The number becomes the first m Number

Program analysis ：
Program source code ：

main()
{
int number[20],n,m,i;
printf("the total numbers is:");
scanf("%d",&n);
printf("back m:");
scanf("%d",&m);
for(i=0;i<n-1;i++)
  scanf("%d,",&number[i]);
scanf("%d",&number[n-1]);
move(number,n,m);
for(i=0;i<n-1;i++)
  printf("%d,",number[i]);
printf("%d",number[n-1]);
}
move(array,n,m)
int n,m,array[20];
{
int *p,array_end;
array_end=*(array+n-1);
for(p=array+n-1;p>array;p--)
  *p=*(p-1);
  *array=array_end;
  m--;
  if(m>0) move(array,n,m);
}

【 Program 69】

subject ： Yes n A circle of individuals , Sequence number . Count from the first person （ from 1 To 3 Number off ）, Where to report 3 Of the people out of the circle , The last one left is the one with the original number .

Program analysis ：
Program source code ：

#define nmax 50
main()
{
int i,k,m,n,num[nmax],*p;
printf("please input the total of numbers:");
scanf("%d",&n);
p=num;
for(i=0;i<n;i++)
  *(p+i)=i+1;
  i=0;
  k=0;
  m=0;
  while(m<n-1)
  {
  if(*(p+i)!=0) k++;
  if(k==3)
  { *(p+i)=0;
  k=0;
  m++;
  }
i++;
if(i==n) i=0;
}
while(*p==0) p++;
printf("%d is left\n",*p);
}

【 Program 70】

subject ： Write a function , Find the length of a string , stay main Input string in function , And output its length .

Program analysis ：
Program source code ：

main()
{
int len;
char *str[20];
printf("please input a string:\n");
scanf("%s",str);
len=length(str);
printf("the string has %d characters.",len);
}
length(p)
char *p;
{
int n;
n=0;
while(*p!='\0')
{
  n++;
  p++;
}
return n;
}

【 Program 71】

subject ： To write input() and output() Function input , Output 5 A student's data record .

Program analysis ：
Program source code ：

#define N 5
struct student
{ char num[6];
  char name[8];
  int score[4];
} stu[N];
input(stu)
struct student stu[];
{ int i,j;
  for(i=0;i<N;i++)
  { printf("\n please input %d of %d\n",i+1,N);
   printf("num: ");
   scanf("%s",stu[i].num);
   printf("name: ");
   scanf("%s",stu[i].name);
    for(j=0;j<3;j++)
    { printf("score %d.",j+1);
     scanf("%d",&stu[i].score[j]);
    }
   printf("\n");
  }
}
print(stu)
struct student stu[];
{ int i,j;
printf("\nNo. Name Sco1 Sco2 Sco3\n");
for(i=0;i<N;i++)
{ printf("%-6s%-10s",stu[i].num,stu[i].name);
  for(j=0;j<3;j++)
   printf("%-8d",stu[i].score[j]);
  printf("\n");
}
}
main()
{
  input();
  print();
}

【 Program 72】

subject ： Create a linked list .

Program analysis ：
Program source code ：

/*creat a list*/
#include "stdlib.h"
#include "stdio.h"
struct list
{ int data;
struct list *next;
};
typedef struct list node;
typedef node *link;
void main()
{ link ptr,head;
int num,i;
ptr=(link)malloc(sizeof(node));
ptr=head;
printf("please input 5 numbers==>\n");
for(i=0;i<=4;i++)
{
  scanf("%d",&num);
  ptr->data=num;
  ptr->next=(link)malloc(sizeof(node));
  if(i==4) ptr->next=NULL;
  else ptr=ptr->next;
}
ptr=head;
while(ptr!=NULL)
{ printf("The value is ==>%d\n",ptr->data);
  ptr=ptr->next;
}
}

【 Program 73】

subject ： Reverse output a list .

Program analysis ：
Program source code ：

/*reverse output a list*/
#include "stdlib.h"
#include "stdio.h"
struct list
{ int data;
  struct list *next;
};
typedef struct list node;
typedef node *link;
void main()
{ link ptr,head,tail; 
  int num,i;
  tail=(link)malloc(sizeof(node));
  tail->next=NULL;
  ptr=tail;
  printf("\nplease input 5 data==>\n");
  for(i=0;i<=4;i++)
  {
   scanf("%d",&num);
   ptr->data=num;
   head=(link)malloc(sizeof(node));
   head->next=ptr;
   ptr=head;
  }
ptr=ptr->next;
while(ptr!=NULL)
{ printf("The value is ==>%d\n",ptr->data);
  ptr=ptr->next;
}}

【 Program 74】

subject ： Connect two linked lists .

Program analysis ：
Program source code ：

#include "stdlib.h"
#include "stdio.h"
struct list
{ int data;
struct list *next;
};
typedef struct list node;
typedef node *link;
link delete_node(link pointer,link tmp)
{if (tmp==NULL) /*delete first node*/
  return pointer->next;
else
{ if(tmp->next->next==NULL)/*delete last node*/
   tmp->next=NULL;
  else /*delete the other node*/
   tmp->next=tmp->next->next;
  return pointer;
}
}
void selection_sort(link pointer,int num)
{ link tmp,btmp;
  int i,min;
  for(i=0;i<num;i++)
  {
  tmp=pointer;
  min=tmp->data;
  btmp=NULL;
  while(tmp->next)
  { if(min>tmp->next->data)
  {min=tmp->next->data;
   btmp=tmp;
  }
  tmp=tmp->next;
  }
printf("\40: %d\n",min);
pointer=delete_node(pointer,btmp);
}
}
link create_list(int array[],int num)
{ link tmp1,tmp2,pointer;
int i;
pointer=(link)malloc(sizeof(node));
pointer->data=array[0];
tmp1=pointer;
for(i=1;i<num;i++)
{ tmp2=(link)malloc(sizeof(node));
  tmp2->next=NULL;
  tmp2->data=array[i];
  tmp1->next=tmp2;
  tmp1=tmp1->next;
}
return pointer;
}
link concatenate(link pointer1,link pointer2)
{ link tmp;
tmp=pointer1;
while(tmp->next)
  tmp=tmp->next;
tmp->next=pointer2;
return pointer1;
}
void main(void)
{ int arr1[]={3,12,8,9,11};
  link ptr;
  ptr=create_list(arr1,5);
  selection_sort(ptr,5);
}

【 Program 75】

subject ： Take it easy , A simple question .

Program analysis ：
Program source code ：

main()
{
int i,n;
for(i=1;i<5;i++)
{ n=0;
  if(i!=1)
  n=n+1;
  if(i==3)
  n=n+1;
  if(i==4)
  n=n+1;
  if(i!=4)
  n=n+1;
  if(n==3)
   printf("zhu hao shi de shi:%c",64+i);
  }
}

【 Program 76】

subject ： Write a function , Input n For even when , Call function to find 1/2+1/4+...+1/n, When the input n In an odd number of , Call function 1/1+1/3+...+1/n( Using pointer functions )

Program analysis ：
Program source code ：

main()
#include "stdio.h"
main()
{
float peven(),podd(),dcall();
float sum;
int n;
while (1)
{
  scanf("%d",&n);
  if(n>1)
   break;
}
if(n%2==0)
{
  printf("Even=");
  sum=dcall(peven,n);
}
else
{
  printf("Odd=");
  sum=dcall(podd,n);
}
printf("%f",sum);
}
float peven(int n)
{
float s;
int i;
s=1;
for(i=2;i<=n;i+=2)
  s+=1/(float)i;
return(s);
}
float podd(n)
int n;
{
float s;
int i;
s=0;
for(i=1;i<=n;i+=2)
  s+=1/(float)i;
return(s);
}
float dcall(fp,n)
float (*fp)();
int n;
{
float s;
s=(*fp)(n);
return(s);
}

【 Program 77】

subject ： Fill in the blanks （ The pointer to the pointer ）

Program analysis ：
Program source code ：

main()
{ char *s[]={"man","woman","girl","boy","sister"};
char **q;
int k;
for(k=0;k<5;k++)
{       ;/* What statement to fill in here */
  printf("%s\n",*q);
}
}

【 Program 78】

subject ： Find the oldest person , And the output . Please find out what's wrong with the program .

Program analysis ：
Program source code ：

#define N 4
#include "stdio.h"
static struct man
{ char name[20];
int age;
} person[N]={"li",18,"wang",19,"zhang",20,"sun",22};
main()
{struct man *q,*p;
int i,m=0;
p=person;
for (i=0;i<N;i++)
{if(m<p->age)
  q=p++;
  m=q->age;}
printf("%s,%d",(*q).name,(*q).age);
}

【 Program 79】

subject ： String sort .

Program analysis ：
Program source code ：

main()
{
char *str1[20],*str2[20],*str3[20];
char swap();
printf("please input three strings\n");
scanf("%s",str1);
scanf("%s",str2);
scanf("%s",str3);
if(strcmp(str1,str2)>0) swap(str1,str2);
if(strcmp(str1,str3)>0) swap(str1,str3);
if(strcmp(str2,str3)>0) swap(str2,str3);
printf("after being sorted\n");
printf("%s\n%s\n%s\n",str1,str2,str3);
}
char swap(p1,p2)
char *p1,*p2;
{
char *p[20];
strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p);
}

【 Program 80】

subject ： There is a pile of peaches on the beach , Five monkeys to share . The first monkey divided the pile of peaches into five parts , One more. , The monkey threw one more into the sea , Took a share of . The second monkey divided the remaining peaches into five parts on average , One more , It also throws one more into the sea , Took a share of , Third 、 Fourth 、 The fifth monkey did this , Ask how many peaches there used to be on the beach ？

Program analysis ：
Program source code ：

main()
{int i,m,j,k,count;
for(i=4;i<10000;i+=4)
{ count=0;
m=i;
for(k=0;k<5;k++)
{
  j=i/4*5+1;
  i=j;
  if(j%4==0)
   count++;
  else
   break;
}
  i=m;
  if(count==4)
  {printf("%d\n",count);
   break;}
}
}

【 Program 81】

subject ：809*??=800*??+9*??+1 among ?? Two digits for ,8*?? It's a double-digit result ,9*?? As the result of the 3 digit . seek ?? Two digits for , And 809*?? After the results of the .

Program analysis ：
Program source code ：

output(long b,long i)
{ printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i);
}
main()
{long int a,b,i;
a=809;
for(i=10;i<100;i++)
{b=i*a+1;
if(b>=1000&&b<=10000&&8*i<100&&9*i>=100)
output(b,i); }
}

【 Program 82】

subject ： Octal to decimal

Program analysis ：
Program source code ：

main()
{ char *p,s[6];int n;
p=s;
gets(p);
n=0;
while(*(p)!='\0')
{n=n*8+*p-'0';
p++;}
printf("%d",n);
}

【 Program 83】

subject ： seek 0—7 The odd number that can be made up of .

Program analysis ：
Program source code ：

main()
{
long sum=4,s=4;
int j;
for(j=2;j<=8;j++)/*j is place of number*/
{ printf("\n%ld",sum);
if(j<=2)
s*=7;
else
s*=8;
sum+=s;}
printf("\nsum=%ld",sum);
}

【 Program 84】

subject ： An even number can always be expressed as the sum of two primes .

Program analysis ：
Program source code ：

#include "stdio.h"
#include "math.h"
main()
{ int a,b,c,d;
scanf("%d",&a);
for(b=3;b<=a/2;b+=2)
{ for(c=2;c<=sqrt(b);c++)
if(b%c==0) break;
if(c>sqrt(b))
d=a-b;
else
break;
for(c=2;c<=sqrt(d);c++)
if(d%c==0) break;
if(c>sqrt(d))
printf("%d=%d+%d\n",a,b,d);
}
}

【 Program 85】

subject ： How many can a prime number be divided into 9 to be divisible by

Program analysis ：
Program source code ：

main()
{ long int m9=9,sum=9;
int zi,n1=1,c9=1;
scanf("%d",&zi);
while(n1!=0)
{ if(!(sum%zi))
n1=0;
else
{m9=m9*10;
sum=sum+m9;
c9++;
}
}
printf("%ld,can be divided by %d \"9\"",sum,c9);
}

【 Program 86】

subject ： Two string linkers

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{char a[]="acegikm";
char b[]="bdfhjlnpq";
char c[80],*p;
int i=0,j=0,k=0;
while(a[i]!='\0'&&b[j]!='\0')
{if (a[i]<b[j])
{ c[k]=a[i];i++;}
else
c[k]=b[j++];
k++;
}
c[k]='\0';
if(a[i]=='\0')
p=b+j;
else
p=a+i;
strcat(c,p);
puts(c);
}

【 Program 87】

subject ： Answer results （ Structure variable transfer ）

Program analysis ：
Program source code ：

#include "stdio.h"
struct student
{ int x;
char c;
} a;
main()
{a.x=3;
a.c='a';
f(a);
printf("%d,%c",a.x,a.c);
}
f(struct student b)
{
b.x=20;
b.c='y';
}

【 Program 88】

subject ： Read 7 Number （1—50） The integer value , Every time a value is read , The program prints out the ＊.

Program analysis ：
Program source code ：

main()
{int i,a,n=1;
while(n<=7)
{ do {
    scanf("%d",&a);
    }while(a<1||a>50);
for(i=1;i<=a;i++)
  printf("*");
printf("\n");
n++;}
getch();
}

【 Program 89】

subject ： A company uses a public telephone to transmit data , The data is a four digit integer , It's encrypted during delivery , The encryption rules are as follows ： Add... To every number 5, Then divide the sum by 10 The remainder of replace the number , Then exchange the first and the fourth , The second and the third exchange .

Program analysis ：
Program source code ：

main()
{int a,i,aa[4],t;
scanf("%d",&a);
aa[0]=a%10;
aa[1]=a%100/10;
aa[2]=a%1000/100;
aa[3]=a/1000;
for(i=0;i<=3;i++)
  {aa[i]+=5;
  aa[i]%=10;
  }
for(i=0;i<=3/2;i++)
  {t=aa[i];
  aa[i]=aa[3-i];
  aa[3-i]=t;
  }
for(i=3;i>=0;i--)
printf("%d",aa[i]);
}

【 Program 90】

subject ： Upgrading from junior college to undergraduate , Read the results .

Program analysis ：
Program source code ：

#include "stdio.h"
#define M 5
main()
{int a[M]={1,2,3,4,5};
int i,j,t;
i=0;j=M-1;
while(i<j)  
{t=*(a+i);
*(a+i)=*(a+j);
*(a+j)=t;
i++;j--;
}
for(i=0;i<m;i++)  
printf("%d",*(a+i));
}

【 Program 91】

subject ： Time function example 1

Program analysis ：
Program source code ：

#include "stdio.h"
#include "time.h"
void main()
{ time_t lt; /*define a longint time varible*/
lt=time(NULL);/*system time and date*/
printf(ctime(<)); /*english format output*/
printf(asctime(localtime(<)));/*tranfer to tm*/
printf(asctime(gmtime(<))); /*tranfer to Greenwich time*/
}

【 Program 92】

subject ： Time function example 2

Program analysis ：
Program source code ：

/*calculate time*/
#include "time.h"
#include "stdio.h"
main()
{ time_t start,end;
int i;
start=time(NULL);
for(i=0;i<3000;i++)
{ printf("\1\1\1\1\1\1\1\1\1\1\n");}
end=time(NULL);
printf("\1: The different is %6.3f\n",difftime(end,start));
}

【 Program 93】

subject ： Time function example 3

Program analysis ：
Program source code ：

/*calculate time*/
#include "time.h"
#include "stdio.h"
main()
{ clock_t start,end;
int i;
double var;
start=clock();
for(i=0;i<10000;i++)
{ printf("\1\1\1\1\1\1\1\1\1\1\n");}
end=clock();
printf("\1: The different is %6.3f\n",(double)(end-start));
}

【 Program 94】

subject ： Time function example 4, A guessing game , Judge how fast a person reacts .（ The moderator made it up when he was a beginner ）

Program analysis ：
Program source code ：

#include "time.h"
#include "stdlib.h"
#include "stdio.h"
main()
{char c;
clock_t start,end;
time_t a,b;
double var;
int i,guess;
srand(time(NULL));
printf("do you want to play it.('y' or 'n') \n");
loop:
while((c=getchar())=='y')
{
i=rand()%100;
printf("\nplease input number you guess:\n");
start=clock();
a=time(NULL);
scanf("%d",&guess);
while(guess!=i)
{if(guess>i)
{printf("please input a little smaller.\n");
scanf("%d",&guess);}
else
{printf("please input a little bigger.\n");
scanf("%d",&guess);}
}
end=clock();
b=time(NULL);
printf("\1: It took you %6.3f seconds\n",var=(double)(end-start)/18.2);
printf("\1: it took you %6.3f seconds\n\n",difftime(b,a));
if(var<15)
printf("\1\1 You are very clever! \1\1\n\n");
else if(var<25)
printf("\1\1 you are normal! \1\1\n\n");
else
printf("\1\1 you are stupid! \1\1\n\n");
printf("\1\1 Congradulations \1\1\n\n");
printf("The number you guess is %d",i);
}
printf("\ndo you want to try it again?(\"yy\".or.\"n\")\n");
if((c=getch())=='y')
goto loop;
}

【 Program 95】

subject ： Family financial management applet

Program analysis ：
Program source code ：

/*money management system*/
#include "stdio.h"
#include "dos.h"
main()
{
FILE *fp;
struct date d;
float sum,chm=0.0;
int len,i,j=0;
int c;
char ch[4]="",ch1[16]="",chtime[12]="",chshop[16],chmoney[8];
pp: clrscr();
sum=0.0;
gotoxy(1,1);printf("|---------------------------------------------------------------------------|");
gotoxy(1,2);printf("| money management system(C1.0) 2000.03 |");
gotoxy(1,3);printf("|---------------------------------------------------------------------------|");
gotoxy(1,4);printf("| -- money records -- | -- today cost list -- |");
gotoxy(1,5);printf("| ------------------------ |-------------------------------------|");
gotoxy(1,6);printf("| date: -------------- | |");
gotoxy(1,7);printf("| | | | |");
gotoxy(1,8);printf("| -------------- | |");
gotoxy(1,9);printf("| thgs: ------------------ | |");
gotoxy(1,10);printf("| | | | |");
gotoxy(1,11);printf("| ------------------ | |");
gotoxy(1,12);printf("| cost: ---------- | |");
gotoxy(1,13);printf("| | | | |");
gotoxy(1,14);printf("| ---------- | |");
gotoxy(1,15);printf("| | |");
gotoxy(1,16);printf("| | |");
gotoxy(1,17);printf("| | |");
gotoxy(1,18);printf("| | |");
gotoxy(1,19);printf("| | |");
gotoxy(1,20);printf("| | |");
gotoxy(1,21);printf("| | |");
gotoxy(1,22);printf("| | |");
gotoxy(1,23);printf("|---------------------------------------------------------------------------|");
i=0;
getdate(&d);
sprintf(chtime,"%4d.%02d.%02d",d.da_year,d.da_mon,d.da_day);
for(;;)
{
gotoxy(3,24);printf(" Tab __browse cost list Esc __quit");
gotoxy(13,10);printf(" ");
gotoxy(13,13);printf(" ");
gotoxy(13,7);printf("%s",chtime);
j=18;
ch[0]=getch();
if(ch[0]==27)
break;
strcpy(chshop,"");
strcpy(chmoney,"");
if(ch[0]==9)
{
mm:i=0;
fp=fopen("home.dat","r+");
gotoxy(3,24);printf(" ");
gotoxy(6,4);printf(" list records ");
gotoxy(1,5);printf("|-------------------------------------|");
gotoxy(41,4);printf(" ");
gotoxy(41,5);printf(" |");
while(fscanf(fp,"%10s%14s%f\n",chtime,chshop,&chm)!=EOF)
{ if(i==36)
{ getch();
i=0;}
if ((i%36)<17)
{ gotoxy(4,6+i);
printf(" ");
gotoxy(4,6+i);}
else
if((i%36)>16)
{ gotoxy(41,4+i-17);
printf(" ");
gotoxy(42,4+i-17);}
i++;
sum=sum+chm;
printf("%10s %-14s %6.1f\n",chtime,chshop,chm);}
gotoxy(1,23);printf("|---------------------------------------------------------------------------|");
gotoxy(1,24);printf("| |");
gotoxy(1,25);printf("|---------------------------------------------------------------------------|");
gotoxy(10,24);printf("total is %8.1f$",sum);
fclose(fp);
gotoxy(49,24);printf("press any key to.....");getch();goto pp;
}
else
{
while(ch[0]!='\r')
{ if(j<10)
{ strncat(chtime,ch,1);
j++;}
if(ch[0]==8)
{
len=strlen(chtime)-1;
if(j>15)
{ len=len+1; j=11;}
strcpy(ch1,"");
j=j-2;
strncat(ch1,chtime,len);
strcpy(chtime,"");
strncat(chtime,ch1,len-1);
gotoxy(13,7);printf(" ");}
gotoxy(13,7);printf("%s",chtime);ch[0]=getch();
if(ch[0]==9)
goto mm;
if(ch[0]==27)
exit(1);
}
gotoxy(3,24);printf(" ");
gotoxy(13,10);
j=0;
ch[0]=getch();
while(ch[0]!='\r')
{ if (j<14)
{ strncat(chshop,ch,1);
j++;}
if(ch[0]==8)
{ len=strlen(chshop)-1;
strcpy(ch1,"");
j=j-2;
strncat(ch1,chshop,len);
strcpy(chshop,"");
strncat(chshop,ch1,len-1);
gotoxy(13,10);printf(" ");}
gotoxy(13,10);printf("%s",chshop);ch[0]=getch();}
gotoxy(13,13);
j=0;
ch[0]=getch();
while(ch[0]!='\r')
{ if (j<6)
{ strncat(chmoney,ch,1);
j++;}
if(ch[0]==8)
{ len=strlen(chmoney)-1;
strcpy(ch1,"");
j=j-2;
strncat(ch1,chmoney,len);
strcpy(chmoney,"");
strncat(chmoney,ch1,len-1);
gotoxy(13,13);printf(" ");}
gotoxy(13,13);printf("%s",chmoney);ch[0]=getch();}
if((strlen(chshop)==0)||(strlen(chmoney)==0))
continue;
if((fp=fopen("home.dat","a+"))!=NULL);
fprintf(fp,"%10s%14s%6s",chtime,chshop,chmoney);
fputc('\n',fp);
fclose(fp);
i++;
gotoxy(41,5+i);
printf("%10s %-14s %-6s",chtime,chshop,chmoney);
}}}

【 Program 96】

subject ： Count the number of substrings in a string

Program analysis ：
Program source code ：

#include "string.h"
#include "stdio.h"
main()
{ char str1[20],str2[20],*p1,*p2;
int sum=0;
printf("please input two strings\n");
scanf("%s%s",str1,str2);
p1=str1;p2=str2;
while(*p1!='\0')
{
if(*p1==*p2)
{while(*p1==*p2&&*p2!='\0')
{p1++;
p2++;}
}
else
p1++;
if(*p2=='\0')
sum++;
p2=str2;
}
printf("%d",sum);
getch();}

【 Program 97】

subject ： Enter some characters from the keyboard , Send them to disk one by one , Until you enter a # until .

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{ FILE *fp;
char ch,filename[10];
scanf("%s",filename);
if((fp=fopen(filename,"w"))==NULL)
{printf("cannot open file\n");
exit(0);}
ch=getchar();
ch=getchar();
while(ch!='#')
{fputc(ch,fp);putchar(ch);
ch=getchar();
}
fclose(fp);
}

【 Program 98】

subject ： Enter a string from the keyboard , Convert all lowercase letters to uppercase letters , Then output to a disk file “test” Kept in .

Enter a string with ！ end .

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{FILE *fp;
char str[100],filename[10];
int i=0;
if((fp=fopen("test","w"))==NULL)
{ printf("cannot open the file\n");
exit(0);}
printf("please input a string:\n");
gets(str);
while(str[i]!='!')
{ if(str[i]>='a'&&str[i]<='z')
str[i]=str[i]-32;
fputc(str[i],fp);
i++;}
fclose(fp);
fp=fopen("test","r");
fgets(str,strlen(str)+1,fp);
printf("%s\n",str);
fclose(fp);
}

【 Program 99】

subject ： There are two disk files A and B, Store one line of letters each , It is required to merge the information in these two documents （ In alphabetical order ）, Output to a new file C in .

Program analysis ：
Program source code ：

#include "stdio.h"
main()
{ FILE *fp;
int i,j,n,ni;
char c[160],t,ch;
if((fp=fopen("A","r"))==NULL)
{printf("file A cannot be opened\n");
exit(0);}
printf("\n A contents are :\n");
for(i=0;(ch=fgetc(fp))!=EOF;i++)
{c[i]=ch;
putchar(c[i]);
}
fclose(fp);
ni=i;
if((fp=fopen("B","r"))==NULL)
{printf("file B cannot be opened\n");
exit(0);}
printf("\n B contents are :\n");
for(i=0;(ch=fgetc(fp))!=EOF;i++)
{c[i]=ch;
putchar(c[i]);
}
fclose(fp);
n=i;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
if(c[i]>c[j])
{t=c[i];c[i]=c[j];c[j]=t;}
printf("\n C file is:\n");
fp=fopen("C","w");
for(i=0;i<n;i++)
{ putc(c[i],fp);
putchar(c[i]);
}
fclose(fp);
}

【 Program 100】

subject ： There are five students , Every student has 3 Results of courses , Enter the above data from the keyboard （ Including the student number , full name , Results of three courses ）, Calculate the grade average , The original data and the calculated average score are stored in the disk file "stud" in .

Program analysis ：
Program source code ：

#include "stdio.h"
struct student
{ char num[6];
char name[8];
int score[3];
float avr;
} stu[5];
main()
{int i,j,sum;
FILE *fp;
/*input*/
for(i=0;i<5;i++)
{ printf("\n please input No. %d score:\n",i);
printf("stuNo:");
scanf("%s",stu[i].num);
printf("name:");
scanf("%s",stu[i].name);
sum=0;
for(j=0;j<3;j++)
{ printf("score %d.",j+1);
scanf("%d",&stu[i].score[j]);
sum+=stu[i].score[j];
}
stu[i].avr=sum/3.0;
}
fp=fopen("stud","w");
for(i=0;i<5;i++)
if(fwrite(&stu[i],sizeof(struct student),1,fp)!=1)
printf("file write error\n");
fclose(fp);
}

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