题目编号：

136

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

参考：

/*
 * @Author: Ray123
 * @Description: 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：45
 * 输出：
 * men = 0, women = 15, child = 30
 * men = 3, women = 10, child = 32
 * men = 6, women = 5, child = 34
 * men = 9, women = 0, child = 36
 * @Date: 2021-05-04 23:11:20
 * @LastEditTime: 2021-05-04 23:49:58
 * @FilePath: \undefinede:\MyApp\Microsoft VS Code\MyAdorableCode\PTA_c\136_BrickMover.c
 */

#include <math.h>
#include <stdio.h>
#include <string.h>


int main()
{
    int n;
    scanf("%d",&n);
    int cnt_m=-1,cnt_w=-1,cnt_c=-1;
    if(n<=2){
        printf("None");
    }else{
        int i=0;
        for(i=n/2;i>=0;i--)
        {
            if((n-3*i)%5==0&&(4*n-2*i)%5==0)
            {
                cnt_m=(n-3*i)/5;
                cnt_w=i;
                cnt_c=(4*n-2*i)/5;
                if(cnt_m>=0&&cnt_c>=0)
                {
                    printf("men = %d, women = %d, child = %d\n",cnt_m,cnt_w,cnt_c);
                }
            
            }
        }
        if(cnt_c==-1&&cnt_m==-1&&cnt_w==-1)
        {
            printf("None");
        }
    

    }
    
  
 
 

    return 0;
}