题干：

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is: 

1) Tom and Jerry come up together with a number K. 

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N. 

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls, …, the Nth tube has exact N pearls. 

4) If Jerry succeeds, he wins the game, otherwise Tom wins. 

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.

Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.

Output

For each game, output a line containing either “Tom” or “Jerry”.

Sample Input

2 
5 1 
1 2 3 4 5 
6 2 
1 2 3 4 5 5 

Sample Output

Jerry 
Tom 

解题报告：

   将每个数和可以组成的数连边，建图。然后跑二分图匹配就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 555 + 5;
int n,m,k;
bool line[2005][2005];
bool used[MAX];
int nxt[MAX],a[MAX];
bool find(int x) {for(int i = 1; i<=n; i++) {if(!used[i] && line[x][i]) {used[i] = 1;if(!nxt[i] || find(nxt[i])) {nxt[i] = x;return 1;}}}return 0 ;
}
int match() {int res = 0;for(int i = 1; i<=n; i++) {memset(used,0,sizeof used);if(find(i)) res++;}return res == n;
}
int main()
{int t;cin>>t;while(t--) {scanf("%d%d",&n,&k);memset(nxt,0,sizeof nxt);memset(line,0,sizeof line);for(int i = 1; i<=n; i++) {scanf("%d",a+i);for(int j = 0; j*k + a[i] <= n; j++) {int now = j*k+a[i];line[i][now] = 1;}}if(match()) puts("Jerry");else puts("Tom");}return 0 ;
}