Game with Pearls

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3685    Accepted Submission(s): 1293

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

 

题意:Jerry和Tom两个人玩游戏,Tom提供N个管子,并且在管子里放珍珠(数量至少一个),并且给出了一个k,现在

Jerry要在这些管子里再放一些珍珠,使得第一个管子里有一个珍珠,第二个管子里有两个珍珠,以此类推,不管之前的顺序怎样,

最后放完后排序之后要符合上面说的规则.

思路:

　　每次排序一次,找到比i小的就加上k,然后再排序.

#include <iostream>
#include <cstring>
#include <algorithm>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;

int a[105];
int main(){
    int n;
    cin>>n;
    while(n--){
        memset(a,0,sizeof(a));
        int m,k;
        cin>>m>>k;
        bool prime = true;
        for(int i=1;i<=m;i++){
            cin>>a[i];
        }
        while(prime){
            sort(a+1,a+1+m);
            bool flag = true;
            for(int i=1;i<=m;i++){
                if(a[i]<i){
                    a[i]+=k;
                    flag = false;
                    break;
                }
                else if(a[i]>i){
                    prime = false;
                    break;
                }
            }
            if(flag) break;
        }
        if(prime){
            cout<<"Jerry"<<endl;
        }else{
            cout<<"Tom"<<endl;
        }
    }
    return 0;
}