C language related programming exercises

1. Program to achieve per 5 All data can be output in a row 3 and 5 to be divisible by ,  But ten people are not 0 Of 3 An integer .

#include<stdio.h> void main( ) { int i,j=0;   for (i=100;i<=1000;i++)      if((i%3==0&&i%5==0)&&((i%100)/10!=0&&(i%100)/10!=3))      {         printf("%5d",i);      j++;      if(j%5==0)         printf("\n");      } }

2. Programming from the keyboard to enter a binary form of string , Convert it to decimal integer output .
 

#include <stdio.h> #include <math.h> #include <string.h> int main() {    char s[21];    long a, i, len, k;    scanf("%s",s);    len = strlen(s);    a = 0; k = 1;    for(i=len-1; i>=0; i--)    {       if(s[i] == '1') a += pow(2.0, len-1-i);    }    printf("%ld\n",a);    return 0; }

3. Give a 100 point mark ,  Output grade is required “A”、“B” 、“C” 、“D”、“E”.90 The above points are “A”, 80~89  It is divided into “B”,70~79 It is divided into “C”,60~69 It is divided into “D”,60 It is divided into the following “E”.

#include<stdio.h> void main() {      float score;      printf("please input a number:\n");      scanf("%f",&score);      if(score>=90)         printf("A\n");      else if(score>=80)         printf("B\n");      else if(score>=70)         printf("C\n");      else if(score>=60)         printf("D\n");      else if(score<60)         printf("E\n"); }

4.  seek 1! +2! +3！ ....+50！（ This programming result is wrong because bytes cannot hold that large number , But there is nothing wrong with the program ）

#include<stdio.h> void main() {  int i,j;  long int t=1,sum=0;  for (i=1;i<=50;i++)  {  t=1;  for(j=1;j<=i;j++)  t=t*j;  sum=sum+t;  }  printf("1!+2!+3!...+50!=%d",sum); }

5   Using bubbling method 10 An integer sort .

#include<stdio.h> void main(  )  {    int i,j,t;    int a[10]={8,9,0,7,6,5,4,3,2,1};   for(i=0;i<9;i++)      for(j=0;j<9-i;j++)       {        if(a[j]>a[j+1])          {           t=a[j];a[j]=a[j+1];a[j+1]=t;        }       }     for(i=0;i<10;i++)     printf("%d",a[i]);   }

6.  seek 500 Prime number within .

#include <stdio.h>  int main() { int a,i;     for (a=2;a<=500;a++)     {         for (i=2;i<=a-1;i++)         {             if (a%i==0)              {                 break;             }         }          if(i>=a-1)             printf("%4d",a);     } }

7. Programming technology 5  A row of data is output by 1、3、5、7 A three digit number consisting of different and non repeating numbers .

#include <stdio.h> void main() {  int a[4]={1,3,5,7};  int i,j,k,n=0;  for(i=0;i<4;i++)  {   for(j=0;j<4;j++)   {    for(k=0;k<4;k++)    {     if(i!=j&&j!=k&&k!=i)   // The three numbers are not equal to each other     {      printf("%3d%d%d",a[i],a[j],a[k]);           n++;         if(n%5==0)                printf("\n");     }    }   }  } }

Add ：1+3+5+……+99=?

#include<stdio.h> void main()  {     int  i=1,sum=0;     while(i<100)        {          sum=sum+i;          i=i+2;         }      printf(“%d”,sum);   }

seek 1×2×3×……×10=?

#include<stdio.h> void main() {      int i,k=1;                for(i=1;i<=10;i++)                       k=k*i;                printf("k=%d",k); }

8. Count all daffodils （ The so-called daffodils number , It's such a three digit number ： This number is equal to the sum of the cubes of its bits . Such as 153=13+53+33 ）

#include<stdio.h> void main() {      int i,gw,sw,bw;         for(i=100;i<1000;i++)         {                gw=i%10;                sw=(i%100)/10;                bw=i/100;                if(gw*gw*gw+sw*sw*sw+bw*bw*bw==i)                       printf("%4d",i);         } }

9.c Language seeking 500 Prime number within , And 10 One line output

#include <stdio.h>  int main() { int a,i,m=0;     for (a=2;a<=500;a++)     {         for (i=2;i<=a-1;i++)         {             if (a%i==0)              {                 break;             }         }                      if(i>=a-1)                      {                             printf("%4d",a);                      m++;                      if(m%10==0)               printf("\n");                      }     } }

10. multiplication table , Align triangle left

#include<stdio.h>        void main()         {int i,j;               for(i=1;i<10;i++)                  { for(j=1; j<=i ;j++)                       printf(“%d*%d=%2d ”,i,j,i*j);                 printf(“\n”);               }        }

11. hold 100～200 Can be 7 Divisible number , Output ten numbers as a line , How many such numbers are finally output .

#Include<stdio.h> main( ) { int n,j=0;    for(n=100;n<=200;n++) {  if (n%7!=0)           continue;           printf("%6d",n);        j++;        if (j%10==0)           printf("\n");       } printf(" \n j=%d\n",j);  }

12. Judge a certain number entered m Prime or not . If it's a prime , Output “YES”, If it is not , Output “NO”.

#include<stdio.h> void main( ) {        int j,m;   printf("Enter an integer number: ");   scanf("%d",&m);   for (j=2; j<=m-1; j++)      if (m%j==0) break;   if (j>=m)     printf("YES\n");   else     printf("NO\n"); }

13. Programming sequence 2～10000 Perfect numbers within .( A factor of a number （ Except for the number itself ） The sum is equal to the number itself .)

for example ：6 What is the best factor 1、2、3, Factor and 1+2+3＝6

therefore 6 It's a complete number

#include<stdio.h> void main( ) {        int i,j,s;        for(i=2;i<=10000;i++)        {                      s=0;               for(j=1;j<=i-1;j++)                      if(i%j==0)                             s+=j;                      if(i==s)                             printf("%6d\n",s);        } }

14. Output 8*8 Black and white chessboard

#include<stdio.h> void main( ) {        int i,j;        for(i=1;i<=8;i++)        {               for(j=1;j<=8;j++)                      if((i+j)%2==0)                             printf("■");                             else                             printf("  ");                             printf("\n");        } }

15. Programming , Output the following graphics . 

*******  *****   ***    *

#include<stdio.h> void main( ) { int i,j;   for (i=1; i<=4; i++)     { for (j=1; j<=i; j++)          printf(" ");        for (j=1;j<=8-(2*i-1);j++)          printf("*");        printf("\n");      }  }

16. Output ad 1 - 5000 year （ contain ） All leap years , And count the total number of leap years ;

#include<stdio.h> void main( ) { int i,j=0;   for (i=1;i<=5000;i++)          if((i%4==0&&i%100!=0)||(i%400==0))                {                 printf("%5d",i);          j++;          }          printf(" Leap year total ：%5d",j); }

17. Enter a positive integer n (1<n<=10), Input again n It's an integer , In reverse order （ Storage ） Output these numbers .
       Such as ： Input n=5; Input again 2,5,8,1,4 Save to array ,
       Then exchange their positions       4,1,8,5,2. Output

#include <stdio.h> void main( ) {        int i, n, temp;        int a[10];        scanf("%d", &n);        for(i=0; i<n; i++)               scanf("%d", &a[i]);        for(i=0; i<n; i++)        {               temp=a[i];               a[i]=a[n-1-i];               a[n-1-i]=temp;        }        for(i=0; i<n; i++)               printf("%d ", a[i]);        printf("\n"); }

18. Program to find a 3x4 The maximum element value of the matrix and its row and column numbers

#include <stdio.h> void main( )  {        int  i,j,row,col,max,a[3][4];   max=a[0][0];   row=0 ;   col=0 ;   for(i=0;i<3;i++)            for(j=0;j<4;j++)             if(a[i][j]>max)            {                  max=a[i][j]; row=i;col=j;           }    printf("row=%d,col=%d",row,col);   }

19. The initial value in a character array “I am happy!” Lowercase characters in become uppercase output .

#include <stdio.h>   void  main( )  { char  c[10]={'i', ' ',’a’…};     int  i;     for(i=0;i<10;i++)         if(c[i]>='a'&&c[i]<='z')c[i]-=32;     for(i=0;i<10;i++)         printf(“%c”,a[i]);      printf(“\n”);   }

20. Receive a string from the keyboard , Find the number of characters in the string .

#include <stdio.h>   void  main( )  {  char  str[30];     int  i=0;     scanf(“%s”,str);     while (str[i] != ‘\0’) i++; printf(“the string lenth is %d”,i);  }