Jobdu-1363-欢乐斗地主[解题代码] | Acm之家


九度-1363-欢乐斗地主[解题代码] | Acm之家
         如果大家玩过欢乐斗地主这个游戏,就一定知道有一个具有"提示"功能的按钮。如果你不知道你现在手里的牌有没有比上家大的牌,并且你也懒得去一张一张地看你手中的牌。这时候你就可以点"提示"按钮,系统会告诉你是否有这样的牌。
         如果你是一个喜欢挑战的人,你就一定会想,能不能写一个程序,让它实现欢乐斗地主中的"提示"按钮的功能。
         现在,我们把"提示"按钮所具有的功能简化,它只需要找出在上家出的牌是"三带一对"的情况下你手中的牌是否比上家的牌大。
输入:
         每组测试数据可能有多组输入,对于每一组输入,
         输入的第一行包括一个整数N(1<=N<=18),代表你手中现在还剩下的扑克牌的张数。
         接下来的一行包括N个数字(1-13,分别代表扑克牌中的A-K),给你的这N个数字是无序的。
         接下来的一行包括五个数字,前三个数字是相同的,后两个数字是相同的,代表上家出的"三带一对"。
输出:
         如果你手中的牌有比上家的"三带一对"大的,输出这样的牌,输出的格式与输入中的第三行相同,即五个数字:前三个是一样的,后两个是一样的,代表你手中的"三带一对"。如果你手中没有比上家的"三带一对"大的牌,请输出"My God"。
04    int n,t;
05    int m,p;//上家 的出的牌
06 
07        while(scanf("%d",&n) != EOF){
08 
09            int arr[16] = {0};
10            for(int i=0; i<n; i++){
11                scanf("%d",&t);
12                if(t==1 || t==2)
13                    t += 13;
14                arr[t] ++;
15            }
16            scanf("%d %d %d %d %d",&m,&m,&m,&p,&p);
17 
18            if(m==1 || m==2)
19                 m += 13;
20            int tag = 0;
21 
22            for(int i=m+1; i<=15; i++){
23                if(tag == 1)
24                    break;
25                if(arr[i] >=3 ){
26                    for(int j=3; j<=15; j++)
27                        if(arr[j] >=2 && i!=j){
28                            tag = 1;
29                            if(i==14 || i==15)
30                                i -= 13;
31                            if(j==14 || j==15)
32                                j -= 13;
33                            printf("%d %d %d %d %d\n",i,i,i,j,j);
34                            break;
35                        }
36                }
37            }
38            if(tag == 0)
39                printf("My God\n");
40        }
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