MOOC Weng Kai C language week 5: 1. cycle control 2. multiple cycles 3. cycle application

  One 、 Cycle control

prime number ： Can only be 1 And the number divided by itself , barring 1.（2、3、5、7、11、13、17、19...）

Write a program ： Read in a number entered by the user x, Then judge whether this number is a prime number .

Ideas ： take except 1 And numbers other than it Except it , If it can be counted by those to be divisible by , Then he's not prime , conversely , Then prime . Because it is increasing , So use for loop （for Cycle fit ： The number is definite or the range is definite ）.  

#include<stdio.h>
int main()
{
    int x;
    scanf("%d",&x);
    int i;
    int isPrime=1;//x Prime number ？
    for(i=2;i<x;i++)
    {
        if(x%i==0)
        {
            isPrime==0;
            break;
        }
    }
    if(isPrime==1)
    {
        printf(" Prime number \n");
    }
      else
    {
      printf(" Not primes \n");
    }
    return 0;
}

break vs continue

break: Out of the loop

continue: Skip this round of the loop and the rest of the statements go to the next round .

Two 、 Multiple cycles

1. Nested loops ： Inside the loop is still a loop .

How to write program output 100 Prime number within ？

Ideas ： You need to construct a loop , Loop can make x from 2 Go to the 100. Use the method of the previous question , Let it enter the circulation body , To determine whether it is a prime number , Then the output . In the circulatory system , If it's a prime , Then the prime number is output , If not, you don't need to output this number .

#include<stdio.h>
int main()
{
    int x;
    for(x=2;x<100;x++)
    {
        int i;
        int isPrime=1;//x Prime number ？
        for(i=2;i<x;i++)
            {
            if(x%i==0)
                {
                    isPrime==0;
                    break;
                }
            }
            if(isPrime==1)
                {
                printf("%d",x);
                }
     }
      printf("\n");
      return 0;
}

Nested loops ： Inside the loop is still a loop .

Before output 50 Prime number ?

Ideas ： Can not use for loop （ Because there is no exact range and number of cycles ）.

You need a counter ：int cnt=0;  Find one prime at a time , The counter needs to be incremented once ：cnt++;    .

Then the cycle condition becomes ： As long as the counter is less than 50, Then continue to cycle ：while(cnt<50)   {  }  （ Is equivalent to for Cycle to while loop , Change a condition ）x The initial value should be 2, And there are x++;       .

use for The cycle can also ：for(x=2;cnt<50;x++)  {  }

#include<stdio.h>
int main()
{
    int x;
    x=2;
    int cnt=0;
    //for(x=2;x<100;x++)
    while(cnt<50)
    {
        int i;
        int isPrime=1;//x Prime number ？
        for(i=2;i<x;i++)
            {
            if(x%i==0)
                {
                    isPrime==0;
                    break;
                }
            }
            if(isPrime==1)
                {
                printf("%d",x);
                cnt ++;
                }
            x++;
     }
      printf("\n");
      return 0;
}

use for The cycle can also ：for(x=2;cnt<50;x++)    {  }

#include<stdio.h>
int main()
{
    int x;
    x=2;
    int cnt=0;
    //for(x=2;x<100;x++)
    //while(cnt<50)
    for(x=2;x<100;x++)
    {
        int i;
        int isPrime=1;//x Prime number ？
        for(i=2;i<x;i++)
            {
            if(x%i==0)
                {
                    isPrime==0;
                    break;
                }
            }
            if(isPrime==1)
                {
                printf("%d",x);
                cnt ++;
                }
            //x++;
     }
      printf("\n");
      return 0;
}

2. Jump out of a nested loop ：break You can only do it for the layer of the loop where it is .

Collect coins ：

How to use 1 horn 、2 Angular sum 5 A dime makes up 10 The amount below yuan ？

The procedure is as follows ：

#include<stdio.h>
int main()
{
	int x;
	int one,two,five;
	//scanf("%d",&x);
	x=2;
	for(one=1;one<x*10;one++){
		for(two=1;two<x*10/2;two++){
			for(five=1;five<x*10/5;five++){
				if(one+two*2+five*5==x*10){
					printf(" It can be used %d individual 1 Angular plus % individual 2 Corner plus 5 Jiao gets yuan \n",one,two,five,x);
                }//if
			}//five
		}//two
	}//one
	return 0;
}

If you want to find the first group that can make up the amount, quit

The idea of the procedure is as follows ：

Wrong situation ①： You can't just add one after a loop statement break, because break You can only do it for the layer of the loop where it is .

Wrong situation ②： You can't simply put break, Then no matter what the situation is, leave this for loop , So this break Always jump out of the floor for loop . Should be , Under certain conditions , When this break occurs , Below break Will happen , Below break Will happen .

So how to do , Under certain conditions , Complete this series of things ？（ use if）.

You need to use a variable ：  ①int exit=0;   ② At the first break Before you do it ：break=1;   ③ second 、 Three break The conditions are ：if(exit==1)break;（ namely ： If exit==1 Words , then break）.

This method is ： The relay break（①： You need a variable to express this thing .②： When we want to break When , Let this variable equal 1.③： Judge whether the variable is equal to 1, To relay break.④： Relay again break, Make it jump out of the loop .）

break and continue： Can only be done on the layer where it is .

Wrong situation ①： You can't just add one after a loop statement break, because break You can only do it for the layer of the loop where it is .

#include<stdio.h>
int main()
{
	int x;
	int one,two,five;
	//scanf("%d",&x);
	x=2;
	for(one=1;one<x*10;one++){
		for(two=1;two<x*10/2;two++){
			for(five=1;five<x*10/5;five++){
				if(one+two*2+five*5==x*10){
					printf(" It can be used %d individual 1 Angular plus % individual 2 Corner plus 5 Jiao gets yuan \n",one,two,five,x);
                    break;//( You can't just add one here break, because break You can only do it for the layer of the loop where it is .)
                }//if
			}//five
		}//two
	}//one
	return 0;
}

Wrong situation ②：  You can't simply put break, Then no matter what the situation is, leave this for loop , So this break Always jump out of the floor for loop . Should be , Under certain conditions , When this break occurs , Below break Will happen , Below break Will happen .

#include<stdio.h>
int main()
{
	int x;
	int one,two,five;
    int exit=0;//*
	//scanf("%d",&x);
	x=2;
	for(one=1;one<x*10;one++){
		for(two=1;two<x*10/2;two++){
			for(five=1;five<x*10/5;five++){
				if(one+two*2+five*5==x*10){
					printf(" It can be used %d individual 1 Angular plus % individual 2 Corner plus 5 Jiao gets yuan \n",one,two,five,x);
                    break;//（ You can't simply add break;）
                }//if
			}//five
            break;//( You can't simply add break;）
		}//two
        break;//（ You can't simply add break;）
	}//one
	return 0;
}

The right way ：

So how to do , Under certain conditions , Complete this series of things ？（ use if）.

You need to use a variable ：  ①int exit=0;   ② At the first break Before you do it ：break=1;   ③ second 、 Three break The conditions are ：if(exit==1)break;（ namely ： If exit==1 Words , then break）.

This method is ： The relay break（①： You need a variable to express this thing .②： When we want to break When , Let this variable equal 1.③： Judge whether the variable is equal to 1, To relay break.④： Relay again break, Make it jump out of the loop .）

#include<stdio.h>
int main()
{
	int x;
	int one,two,five;
    int exit=0;//*
	//scanf("%d",&x);
	x=2;
	for(one=1;one<x*10;one++){
		for(two=1;two<x*10/2;two++){
			for(five=1;five<x*10/5;five++){
				if(one+two*2+five*5==x*10){
					printf(" It can be used %d individual 1 Angular plus % individual 2 Corner plus 5 Jiao gets yuan \n",one,two,five,x);
                    //（ You can't just add one here break, because break You can only do it for the layer of the loop where it is .）
                    //（ You can't simply add break;）
                    exit=1;//*
                    break;
                }//if
			}//five
            //( You can't simply add break;）  
            if(exit==1)break;//*
		}//two
        //（ You can't simply add break;）
        if(exit==1)break;//*
	}//one
	return 0;
}

  Method 2：

Of course , You can also use goto The way

goto out;       out:

Where you need to leave the loop , Put in goto sentence .、

goto A label is required after the statement , This label is written in the program itself ,（ Such as goto out;     out：      ）goto To jump to out Go to the place indicated .

In need of multi-layer circulation 、 Inner layer of nested loop , When you jump out , have access to goto.

#include<stdio.h>
int main()
{
	int x;
	int one,two,five;
	scanf("%d",&x);
	for(one=1;one<x*10;one++){
		for(two=1;two<x*10/2;two++){
			for(five=1;five<x*10/5;five++){
				if(one+two*2+five*5==x*10){
					printf(" It can be used %d individual 1 Angular plus % individual 2 Corner plus 5 Jiao gets yuan \n",one,two,five,x);
				    goto out;//*
                }//if
			}//five
		}//two
	}//one
out://*
	return 0;
}

3、 ... and 、 Recycling  

1. Sum up

/1: seek f(n)=1+½+⅓+¼+...+1/n

Ideas ： use for loop （ Because the starting point 1, End n, The scope is clear ） Note floating point numbers .

#include<stdio.h>
int main()
{
int n;
int i;
double sum=0.0;// 1 Divide i The value of has a decimal point 
scanf("%d",&n);
for(i=1;i<=n;i++)
{
    sum+=1.0/i;// 1 Divide i,i Certain ratio 1 Big , There is a floating point number on both sides of the division sign , Then the other will be floating point numbers , The result will be the operation result of floating-point numbers .
}
printf("f(%d)=%f\n",n,sum);
return 0;
}

/2: seek f(n)=1-½+⅓-¼+...+1/n

Ideas ： It's the same idea as the previous one , use for loop （ Because the starting point 1, End n, The scope is clear ） Note floating point numbers . Just one more sign change , You can set another variable , To make changes in symbols .

#include<stdio.h>
int main()
{
int n;
int i;
double sum=0.0;
int sign=1;// Method 2：double sign=1.0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
    sum+=sign*1.0/i;// Method 2:sum+=sign/i;
    sign=-sign;// Method 2:sign=-sign; It is equivalent to using molecules to change positive and negative .
}
printf("f(%d)=%f\n",n,sum);
return 0;
}

2. Find the greatest common divisor

Enter two numbers a and b. Output their greatest common divisor ： Input 12 18; Output ：6.

Ideas ：1. Enumeration （ The efficiency is not high ）   2. division  

1） Enumeration  

int a,b;
int min;
scanf("%d %d",&a,&b);
if(a<b)
    {min=0;}
else
    {min=b;}
int ret=0;
int i;
for(i=1;i<min;i++)
    {if(a%i==0){
        if(b%i==0){
            ret=i;
        }
    }
}
printf("%d and %d The greatest common divisor of %d.\n",a,b,ret);

2） division  

1) If b be equal to 0, End of calculation ,a Is the greatest common divisor ;

2) otherwise , Calculation a Divide b The remainder of , Give Way a be equal to b, and b Equal to the remainder ;

3) Back to step one

If you ask 12 And 18 Maximum common divisor of ： Input 12 18; Output ：6.

a          b          t( remainder )

12       18         12

18       12          6

12        6           0

 6         0

int a,b;
int t;
scanf("%d %d",&a,&b);
while(b!=0)
{
t=a%b;
a=b;
b=t;
}
printf("gcd=%d\n",a);

3. Integer decomposition  

Positive order factorization of integers

Enter a non negative integer , Positive sequence output every digit of it . Input ：13425; Output ：1 3 4 2 5.

int x;
scanf("%d",&x);
int mask=1;
int t=x;
while(t>9){
    t/=10;
    mask*=10;
}
printf("x=%d,mask=%d\n",x,mask);
do{
    int d=x/mask;
    printf("%d",d);
    if(mask>9){
        printf(" ");
    }
    x%=mask;
    mask/=10;
    }while(mask>0);
    printf("\n");

（ This is a bit of a detour for me T^T）