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）

 

题意：给你n个数，(1<=n<=100),还有一个k(1<=k<=n)，然后允许给某一个数加上k的正整数倍，当然可以不加，问你是否可以把这n个数变成1,2,3,...,n,可以就输出Jerry,否则输出Tom。

分析：一开始直接开个100数组去记录每个数出现的次数。。。如果某一个数出现了，那就不用动这个数了，对于没出现的数，枚举要加的数(k,2k,3k,...注意边界，不能超过这个数)。如果存在某一个数没出现而且不能通过别的数加若干个k,直接判断不能实现 。。简简单单就过了

 

来源：http://mathlover.info/archives/hdu5090

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int main()
{
    int m,n,k;
    cin>>m;
    int a[105],cnt[102];
    bool p[105];
    while(m--)
    {
        cin>>n>>k;
        memset(cnt,0,sizeof(cnt));
        memset(p,0,sizeof(p));

        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(p[a[i]]==0)///判断出现的标志
                p[a[i]]=1;
            else
                cnt[a[i]]++;///多余的数字
        }

        bool flag=1;///标记谁能赢
        for(int i=1;i<=n;i++)
        {
            if(p[i]==0)///如果这个数没有出现
            {
                bool ok=0;
                for(int j=1;i-j*k>=1;j++)///依次减去k的倍数
                {
                    if(cnt[i-j*k])///若多余
                    {
                        cnt[i-j*k]--;
                        ok=1;
                        break;///即是可以由多余的数字变化而来，逆过程
                    }
                }
                if(!ok)///ok==0-------若果没有找到，则进行此if
                {
                    flag=0;
                    break;
                }
            }
        }

        if(flag)
            puts("Jerry");
        else
            puts("Tom");
    }
    return 0;
}