换分币问题.c

//将5元的人民币兑换成1元、5角和1角的硬币，共有多少种不同的兑换方法。

#include<stdio.h>
int main()
{
    int x, y, z, count=1;
    printf("可能的兑换方法如下：\n");
    for( x=0; x<=50; x+=10 )  /*x为1元硬币钱数，其取值为0,10,20,30,40,50*/
        for( y=0; y<=50-x; y+=5 )  /*y为5角硬币钱数，其取值为0,5,10,15,20,25,30,35,40,,45,50*/
            for( z=0; z<=50-x-y; z++)  /*z为1角硬币钱数，其取值为0,1,...50*/
                if(x+y+z==50)
                {
                    /*输出时，每行最多三种情况*/
                    printf(count%3 ? "%d: 10*%d+5*%d+1*%d\t" : "%d:10*%d+5*%d+1*%d\n", count, x/10, y/5, z);
                    count++;
                }
    return 0;
}

//运行结果：

//可能的兑换方法如下：
//1: 10*0+5*0+1*50        2: 10*0+5*1+1*45        3:10*0+5*2+1*40
//4: 10*0+5*3+1*35        5: 10*0+5*4+1*30        6:10*0+5*5+1*25
//7: 10*0+5*6+1*20        8: 10*0+5*7+1*15        9:10*0+5*8+1*10
//10: 10*0+5*9+1*5        11: 10*0+5*10+1*0       12:10*1+5*0+1*40
//13: 10*1+5*1+1*35       14: 10*1+5*2+1*30       15:10*1+5*3+1*25
//16: 10*1+5*4+1*20       17: 10*1+5*5+1*15       18:10*1+5*6+1*10
//19: 10*1+5*7+1*5        20: 10*1+5*8+1*0        21:10*2+5*0+1*30
//22: 10*2+5*1+1*25       23: 10*2+5*2+1*20       24:10*2+5*3+1*15
//25: 10*2+5*4+1*10       26: 10*2+5*5+1*5        27:10*2+5*6+1*0
//28: 10*3+5*0+1*20       29: 10*3+5*1+1*15       30:10*3+5*2+1*10
//31: 10*3+5*3+1*5        32: 10*3+5*4+1*0        33:10*4+5*0+1*10
//34: 10*4+5*1+1*5        35: 10*4+5*2+1*0        36:10*5+5*0+1*0