C: 第三种循环(笔记)

int n;

scanf("%d",&n);
int fact = 1;

int i = 1;
while ( i <= n ){
	fact *= i;
	i++;
}

printf("%d!=%d\n",n,fact);

用for循环表示：

for循环后大括号内三个条件（初始动作；循环继续的条件；循环每轮要做的动作）

int n;

scanf("%d",&n);
int fact = 1;

int i = 1;
for ( i=1; i<=n; i++ ){
	fact *= i;
}

printf("%d!=%d\n",n,fact);

 

---

离开多重循环：

例子  凑硬币

#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("可以用%d个1角加%d个5角得到%d元\n",
						one, two, five, x);
				}
			}
		}
	}
	
	return 0;
}

加入break

若要出现一种可能改为：

 #include <stdio.h>

int main()
{
	int x;
	int one,two,five;
	int exit = 0;
	
	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("可以用%d个1角加%d个5角得到%d元\n",
						one, two, five, x);
					
					exit = 1;
					break;
				}
			}
			if ( exit ==1) break;
		}
		if ( exit ==1 ) break;
	}
	
	return 0;
}

也可运用goto out;    out: 

#include <stdio.h>
		
int main()
{
	int x;
	int one,two,five;
	int exit = 0;
			
	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("可以用%d个1角加%d个5角得到%d元\n",
						one, two, five, x);
					goto out;
				}
			}
		}
	}
out:	
	return 0;
}

---

 100以内的素数

#include <stdio.h>

int main()
{
	int x;
	
	for ( x=1; x<=100; x++){
		int i;
		int isPrime = 1;  // x是素数
		for ( i=2;i<x;i++){
			if( x % i == 0){
				isPrime = 0;
				break;
			}
		}
		if ( isPrime == 1){
			printf("%d ",x);
		}
	}
	return 0;
}

50个素数

#include <stdio.h>

int main()
{
	int x;
	int cnt = 0;//
	
	for ( x=1; cnt<50; x++){
		int i;
		int isPrime = 1;  // x是素数
		for ( i=2;i<x;i++){
			if( x % i == 0){
				isPrime = 0;
				break;
			}
		}
		if ( isPrime == 1){
			cnt++;
			printf("%d\t ",x);   //加个\t让数字对齐
			if ( cnt %5 == 0){   //令50个素数排成5个一行
				printf("\n");
			}
		}
	}
	return 0;
}

---

sum=1+1/2+1/3+......+1/n求和

#include <stdio.h>

int main()
{
	int n;
	int i;
	double sum = 0.0;
	
	scanf("%d",&n);
	for ( i = 1; i<=n; i++){
		sum += 1.0/i;
	}
	
	printf("f(%d)=%f\n",n,sum);
	
	return 0;
}

+-+-+-+-

#include <stdio.h>

int main()
{
	int n;
	int i;
	double sum = 0.0;
	int sign = 1;
	
	scanf("%d",&n);
	for ( i = 1; i<=n; i++){
		sum += sign*1.0/i;
		sign = -sign;
	}
	
	printf("f(%d)=%f\n",n,sum);
	
	return 0;
}

---

求最大公约数

1. 枚举法
2. 辗转相除法：

---

正序分解整数