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。。。

View Code 
 1 #include<stdio.h>
 2 int main(){
 3     int  sum,x1,x5,x10;
 4     int  a,b,c;
 5     while( scanf("%d%d%d%d",&sum,&x1,&x5,&x10)!=EOF && ( sum||x1||x5||x10) ){
 6         a = b = c = 0;
 7         if( (sum%5)>x1 ){
 8             printf("Hat cannot buy tea.\n");
 9             continue;
10         }    
11         a = sum%5;
12         sum -= a;
13         x1 -= a;
14         if( sum<=x1 ){
15             a += sum;
16             printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
17             continue;
18         }
19         x1/=5;
20         sum -= x1*5;
21         a += 5*x1;
22         if( sum<=5*x5 ){
23             b += sum/5;
24             printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
25             continue;
26         }
27         sum -= 5*x5;
28         b = x5;
29         if( sum%10==0 ){
30             if( sum<=10*x10 ){
31                 c += sum/10;
32                 printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
33                 continue;
34             }
35             else{
36                 printf("Hat cannot buy tea.\n");
37                 continue;
38             }
39         }
40         else{
41             if( b>0 )
42                 b--;
43             else
44                 if( a>=5 )
45                     a-=5;
46                 else{
47                      printf("Hat cannot buy tea.\n");
48                      continue;
49                 }
50             sum+=5;
51             if( sum<=10*x10 ){
52                 c = sum/10;
53                 printf("%d YiJiao, %d WuJiao, and %d ShiJiao\n",a,b,c);
54                 continue;
55             }
56             else{
57                 printf("Hat cannot buy tea.\n");
58                 continue;
59             }
60         }
61     }
62     return 0;
63 }

 

posted @ 2013-05-05 19:47  xxx0624  阅读(308)  评论(0)  编辑  收藏  举报