D.Game with Pearls

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

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

N个盘子，每个盘子可以增加0或k的正整数倍，问最终能否形成一个1-N的序列，

寻找n次依次找没用过的且可以使它符合要求的盘子，全找到即为Jerry赢。

#include<bits/stdc++.h>
#include<stdio.h>
using namespace std;

int main()
{
    int m;
    cin >> m;
    while(m--)
    {
        int n,k;
        cin >> n >> k;
        int a[105];
        int temp[105];
        memset(a, 0, sizeof(a));
        memset(temp, 0, sizeof(temp));
        for(int i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        sort(a + 1, a + n + 1);
        int flag;//标记能否满足第i个tube里有i个pearl
        for(int i = 1; i <= n; i++)
        {
            flag = 0;
            for(int j = 1; j <= n; j++)
            {
                if((!temp[j]) && (i >= a[j]) && ((i - a[j]) % k == 0))//之前没装过可以装pearls
                {
                    temp[j] = 1;//标记第i个tube已经装过
                    flag = 1;
                    break;
                }
            }
            if(!flag)//找了一圈仍没找到~失败
            {
                break;
            }
        }
        if(flag){
            printf("Jerry\n");
        }
        else{
            printf("Tom\n");
        }
    }
    return 0;
}