5重循环如何赋值？拿50分谢谢啦

[qijixingzhe]

#include "stdafx.h"
#include<stdio.h>
#include<math.h>
int main()
{
int a, b, c, d, e, i = 0;
int sum = 0;
int b1[31] = { 1,4,3,4,5,6,7,0 };
int b2[31] = { 1,1,3,4,5,3,7,0 };
int b3[31] = { 1,7,3,4,5,6,7,0 };
int b4[31] = { 1,1,3,4,5,4,7,0 };
int b5[31] = { 1,1,3,4,5,6,7,0 };
for (a = 1; a <= 8; a++)
{
b1[a] = 1;
b1[a] = 4;
b1[a] = 3;
b1[a] = 4;
b1[a] = 5;
b1[a] = 6;
b1[a] = 7;
b1[a] = 0;
}

for (b = 1; b <= 8; b++)
{
b2[b] = 1;
b2[b] = 1;
b2[b] = 3;
b2[b] = 4;
b2[b] = 5;
b2[b] = 3;
b2[b] = 7;
b2[b] = 0;
}
for (c = 1; c <= 8; c++)
{
b3[c] = 1;
b3[c] = 7;
b3[c] = 3;
b3[c] = 4;
b3[c] = 5;
b3[c] = 6;
b3[c] = 7;
b3[c] = 0;
}
for (d = 1; d <= 8; d++)
{
b4[d] = 1;
b4[d] = 1;
b4[d] = 3;
b4[d] = 4;
b4[d] = 5;
b4[d] = 4;
b4[d] = 7;
b4[d] = 0;
}
for (e = 1; e <= 8; e++)
{
b5[e] = 1;
b5[e] = 1;
b5[e] = 3;
b5[e] = 4;
b5[e] = 5;
b5[e] = 6;
b5[e] = 7;
b5[e] = 0;
}

sum = b1[a] + b2[b] + b3[c] + b4[d] + b5[e];
if (sum ==5)
{
printf("A:%2d B:%2d C:%2d D:%2d E:%2d ", b1[a], b2[b], b3[c], b4[d], b5[e]);
i++;
if (i % 1 == 0)
printf("\n");
}
printf("共有%d种办法\n", i);
return 0;
}
我想当a=1的时候b1【1】=1，依次类推，最后判定sum==5把它们分别打印出来，但是运行结果却现实共有0种办法，应该最少有一种组合的呀。问题出在哪里啦？5重for循环，如何用数组给它分别循环赋值呢？请教

[考拉一枚]

你知道你这里计算的是什么 sum = b1[a] + b2[b] + b3[c] + b4[d] + b5[e]; 在你的代码里这里算的是 sum = b1[9] + b2[9] + b3[9] + b4[9] + b5[9]=0，接着if循环判断为假，继续执行printf("共有%d种办法\n", i); 这里的i值为0.。。。。

[L-talent]

数组赋值:如果是常量的话直接赋值就好了，而且你已经赋过了。非要用循环的话，要是变量。比如 for(int i=0;i<8;i++) { b1[i]=变量; }

[@风轻云淡_]

引用 5 楼 qq_35876885 的回复:

数组赋值:如果是常量的话直接赋值就好了，而且你已经赋过了。非要用循环的话，要是变量。比如 for(int i=0;i<8;i++) { b1[i]=变量; }

很明显，他要得是每个数组元素里面有不同的值，怎么用循环赋值？

[LubinLew]

#include "stdafx.h"
#include<stdio.h>

#define MAX 8

int main(void)
{
	int a, b, c, d, e, i = 0;
	int sum = 0;
	int b1[MAX] = { 1,4,3,4,5,6,7,0 };
	int b2[MAX] = { 1,1,3,4,5,3,7,0 };
	int b3[MAX] = { 1,7,3,4,5,6,7,0 };
	int b4[MAX] = { 1,1,3,4,5,4,7,0 };
	int b5[MAX] = { 1,1,3,4,5,6,7,0 };

	for (a = 0; a < MAX; a++) {
		for (b = 0; b < MAX; b++) {
			for (c = 0; c < MAX; c++) {
				for (d = 0; d < MAX; d++) {
					for (e = 0; e < MAX; e++) {
						sum = b1[a] + b2[b] + b3[c] + b4[d] + b5[e];
						if (5 == sum) {
							printf("A:%2d B:%2d C:%2d D:%2d E:%2d ", b1[a], b2[b], b3[c], b4[d], b5[e]);
							i++;
							if (i % 1 == 0)
								printf("\n");
						}
					}
				}
			}
		}
	}

	printf("共有%d种办法\n", i);

return 0;
}

[L-talent]

比如int b1[9]; int a[9]={1,3,5,7,9,0,2,4,6}; for(int i=0;i<9;i++) { b1[i]=a[i]; } 不过一般a[9]是其他表达式或函数获得

[自信男孩]

for (a = 1; a <= 8; a++)
{
b1[a] = 1;
b1[a] = 4;
b1[a] = 3;
b1[a] = 4;
b1[a] = 5;
b1[a] = 6;
b1[a] = 7;
b1[a] = 0;
}

这个循环有很多冗余的操作 等价与下面的

for (a = 1; a <= 8; a++)
{
b1[a] = 0;
}

所以要么你的循环写的有问题，要么就是逻辑考虑的有问题 下面的几个循环都是这样的问题

[自信男孩]

sum = b1[a] + b2[b] + b3[c] + b4[d] + b5[e];

这儿的sum的值是0，因为b1[a]。。。b5[e]都是0。因为下标是从0开始，而不是1开始。 所以上面的sum = b1[a] + b2[b] + b3[c] + b4[d] + b5[e];等价sum = b1[9] + b2[9] + ... + b5[9]; b1[9] ... b5[9]都是0。

[@风轻云淡_]

你定义数组的时候就初始化了，假设后面的所谓的循环正确的话 他所做的也就是你已经做过的初始化工作，反而现在你的循环将所有元素都变为了0 你后面又判断他们等不等于5，全是0 ，哪来的5，好好理顺一下逻辑吧。