Question

c语言这到程序怎么写，第四题

Answer 1

#include <stdio.h>
int main()
{
	int pi,xi,li,pc,xc,lc,n;
	pc=4;
	xc=40;
	lc=2;
	n=0;
	for(pi=1;pi<100;++pi){
		for(xi=1;xi<10;++xi){
			for(li=1;li<200;++li){
				if((pi+xi+li)==100){
					if((pi*4+xi*40+li*2)==400){
						++n;
						printf("方案%2d: 苹果%2d个 西瓜%2d个 梨%2d个\n",n,pi,xi,li);
					}
				}
			}
		}
	}
	printf("共有%d个方案\n",n);
	
	getchar();
	return 0;
}

Answer 2

#include <stdio.h>
int main()
{
	int apple,pear,watermelon;
	int count=0;
	for(apple=1;apple<100;apple++)
		for(pear=1;pear<100;pear++)
			for(watermelon=1;watermelon<10;watermelon++)
			{
				if(apple+pear+watermelon==100 && 2*apple+pear+20*watermelon==200)
				{
					printf("apple=%d pear=%d watermelon=%d\n",apple,pear,watermelon);
					count++;
				}
			}
	printf("购买方案总共有: %d\n",count);
	return 0;
}

运行结果如下：

apple=5 pear=90 watermelon=5

apple=24 pear=72 watermelon=4

apple=43 pear=54 watermelon=3

apple=62 pear=36 watermelon=2

apple=81 pear=18 watermelon=1

购买方案总共有: 5