HDU5090——贪心——Game with Pearls

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

 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

 

 大意：给出n个管子，以及管子里面珍珠的数目，问你向里面假如0或者k的倍数，是否能够满足i = a[i]，如果都满足输出Jerry，没有考虑其他也可以转变成这个并不是一一对应的关系

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[150];
int b[150];
int main()
{
    int T;
    int n,k;
    scanf("%d",&T);
    while(T--){
        memset(b,0,sizeof(b));
        scanf("%d%d",&n,&k);
        for(int i = 1; i <= n ; i++)
            scanf("%d",&a[i]);
        for(int i = 1; i <= n ; i++){
            for(int p = a[i]; p <= n ; p+=k){
                b[p]++;
            }
        }
        int flag = 1;
        for(int i = 1; i <= n && flag ; i++){
            if(!b[i]){
                flag = 0;
                break;
            }
            else {
                for(int j = i; j <= n ; j+=k)
                    b[j] --;
            }
        }
        if(flag)
            printf("Jerry
");
        else printf("Tom
");
    }
    return 0;
}