HDU1288+枚举

虽然题意简单，但是注意的细节还是不少。

1、必须满足能刚好付这么多钱。

2、尽可能先把1毛的用完

3、在2的基础上把5毛的用完

4、用10的时候，保证sum是10的倍数，否则先去掉5个1毛，没有1毛再去5毛。。。

#include<stdio.h>
int main(){
    int  sum,x1,x5,x10;
    int  a,b,c;
    while( scanf("%d%d%d%d",&sum,&x1,&x5,&x10)!=EOF && ( sum||x1||x5||x10) ){
        a = b = c = 0;
        if( (sum%5)>x1 ){
            printf("Hat cannot buy tea.\n");
            continue;
        }    
        a = sum%5;
        sum -= a;
        x1 -= a;
        if( sum<=x1 ){
            a += sum;
            printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
            continue;
        }
        x1/=5;
        sum -= x1*5;
        a += 5*x1;
        if( sum<=5*x5 ){
            b += sum/5;
            printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
            continue;
        }
        sum -= 5*x5;
        b = x5;
        if( sum%10==0 ){
            if( sum<=10*x10 ){
                c += sum/10;
                printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
                continue;
            }
            else{
                printf("Hat cannot buy tea.\n");
                continue;
            }
        }
        else{
            if( b>0 )
                b--;
            else
                if( a>=5 )
                    a-=5;
                else{
                     printf("Hat cannot buy tea.\n");
                     continue;
                }
            sum+=5;
            if( sum<=10*x10 ){
                c = sum/10;
                printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
                continue;
            }
            else{
                printf("Hat cannot buy tea.\n");
                continue;
            }
        }
    }
    return 0;
}