ZOJ 3610Yet Another Story of Rock-paper-scissors

Akihisa and Hideyoshi were lovers. They were sentenced to death by the FFF Inquisition. Ryou, the leader of the FFF Inquisition, promised that the winner of Rock-paper-scissors would be immune from the punishment. Being lovers, Akihisa and Hideyoshi decided to die together with their fists clenched, which indicated rocks in the game. However, at the last moment, Akihisa chose paper and Hideyoshi chose scissors. As a result, Akihisa was punished by the FFF Inquisition and Hideyoshi survived alone.

When a boy named b and a girl named g are being punished by the FFF Inquisition, they will play Rock-paper-scissors and the winner will survive. If there is a tie, then neither of they will survive. At first, they promise to choose the same gesture x. But actually, the boy wants to win and the girl wants to lose. Of course, neither of them knows that the other one may change his/her gesture. At last, who will survive?

Input

There are multiple test cases. The first line of input is an integer T ≈ 1000 indicating the number of test cases.

Each test case contains three strings -- b g x. All strings consist of letters and their lengths never exceed 20. The gesture x is always one of "rock", "paper" and "scissors".

Output

If there is a tie, output "Nobody will survive". Otherwise, output "y will survive" where y is the name of the winner.

Sample Input

1Akihisa Hideyoshi rock

Sample Output

Hideyoshi will survive

简单题

#include<map>#include<cmath>    #include<queue>    #include<vector>#include<cstdio>    #include<cstring>    #include<algorithm>    using namespace std;#define ms(x,y) memset(x,y,sizeof(x))    #define rep(i,j,k) for(int i=j;i<=k;i++)    #define per(i,j,k) for(int i=j;i>=k;i--)    #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    #define inone(x) scanf("%d",&x)    #define intwo(x,y) scanf("%d%d",&x,&y)    #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)    typedef long long LL;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e3 + 10;int T;char x[N], y[N], z[N];int main(){for (inone(T); T--;){scanf("%s%s%s", x, y, z);printf("%s will survive\n", y);}return 0;}