Game with Pearls

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 

题意：Jerry 和 Tom 玩一个游戏 ， 给你 n 个盒子 ， a[ i ] 表示开始时 ，第 i 个盒子中的小球的个数 。 然后 Jerry 可以在每个
盒子里加入 0 或 k的倍数的小球 ， 操作完后，Jerry 可以重新排列 盒子的顺序，最终使 第 i 个盒子中有 i 个小球。 若Jerry能
使最终的盒子变成那样，就输出 “Jerry” ，否则 输出 “Tom” 。

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int a[1000],b[1000];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        memset(b,0,sizeof(b));        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        int sum;        int sum1;        int p=0;        for(int i=1;i<=n;i++)        {            sum1=0;            sum=0;            for(int j=1;j<=n;j++)            {                if((i-a[j])%m==0&&(i-a[j])>=0&&a[j]!=-1&&sum==0)                {                    sum=1;                    a[j]=-1;                }            }            if(sum==1)            {                continue;            }            else            {                p=1;            }        }        if(p==0)        {            printf("Jerry\n");        }        else        {            printf("Tom\n");        }    }}