A - Game with Pearls（最大匹配）

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：

输入t 组样例

在输入n和k

n代表有n个管子下一行代表每个管子初始有几个球可往每个管子里放入k的倍数个球或者不放

如果对应第i个管子里有i个球那么jerry胜利否则tom胜利

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思路：比如第二组样例输入一个数就把对应的check赋1 表示已经满足第i个管子里有i个球

最后一个5在对应的a[5]++表示有一个多的5

然后找check里为0的数用他不断减k看是否有a里面有的数

代码：

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[105];
int check[105];
int main()
{
    int m,n,k,i,flag,b;
    cin>>m;
    while(m--)
    {
        cin>>n>>k;
        flag=0;
        memset(a,0,sizeof(a));
        memset(check,0,sizeof(check));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&b);
            if(check[b]==0)
                check[b]=1;    //表示已满足的
            else
                a[b]++;        //表示多余的管子

        }
        for(int i=1;i<=n;i++)   //1-n个管子里不满足第i个管子有i个球的
        {
           if(check[i]==0)
           {
               int ok=0;     //ok是一个判断标志
               for(int j=1;i-j*k>=1;j++)     //每次用i减去k的倍数
               {
                   if(a[i-j*k]>0)           //如果a里有 ok标志为1
                   {
                       ok=1;
                       a[i-j*k]--;          //a里减去对应的个数
                       check[i]=1;           //i标记为满足
                       break;
                   }
               }
               if(ok==0)                //表示不成功 break掉
               {
                   flag=1;             
                   break;
               }
           }
        }



        if(flag==0)
            printf("Jerry\n");
        else
            printf("Tom\n");
    }
    return 0;
}

A - Game with Pearls（最大匹配）

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