PTA-7-2 穷举问题-搬砖 (15分)

7-2 穷举问题-搬砖 (15分)
某工地需要搬运砖块，已知男人一人搬3块，女人一人搬2块，小孩两人搬1块。如果想用n人正好搬n块砖，问有多少种搬法？

输入格式:
输入在一行中给出一个正整数n。

输出格式:
输出在每一行显示一种方案，按照"men = cnt_m, women = cnt_w, child = cnt_c"的格式，输出男人的数量cnt_m，女人的数量cnt_w，小孩的数量cnt_c。请注意，等号的两侧各有一个空格，逗号的后面也有一个空格。

如果找不到符合条件的方案，则输出"None"

输入样例:
45

输出样例:
men = 0, women = 15, child = 30
men = 3, women = 10, child = 32
men = 6, women = 5, child = 34
men = 9, women = 0, child = 36
某工地需要搬运砖块，已知男人一人搬3块，女人一人搬2块，小孩两人搬1块。如果想用n人正好搬n块砖，问有多少种搬法？

输入格式:
输入在一行中给出一个正整数n。

输出格式:
输出在每一行显示一种方案，按照"men = cnt_m, women = cnt_w, child = cnt_c"的格式，输出男人的数量cnt_m，女人的数量cnt_w，小孩的数量cnt_c。请注意，等号的两侧各有一个空格，逗号的后面也有一个空格。

如果找不到符合条件的方案，则输出"None"

输入样例:
45

输出样例:
men = 0, women = 15, child = 30
men = 3, women = 10, child = 32
men = 6, women = 5, child = 34
men = 9, women = 0, child = 36

#include
int main()
{
	int n;
	scanf("%d", &n);
	int cnt_m = 0;
	int cnt_w = 0;
	float cnt_c = 0;
	int count = 0;
	for (cnt_m = 0; cnt_m <= n; cnt_m++)
	{
		for (cnt_w = 0; cnt_w <= n; cnt_w++)
		{
			for (cnt_c = 0; cnt_c <= n; cnt_c++)
			{
				if ((cnt_m + cnt_w + cnt_c == n) && (cnt_m * 3 + cnt_w * 2 + cnt_c*0.5 == n))
				{
					count++;
					printf("men = %d, women = %d, child = %.0f\n", cnt_m, cnt_w, cnt_c);
				}
			}
		}
	}
	if (count == 0)
	{
		printf("None");
	}
}