Game with Pearls
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
一个博弈题，因为数据量比较小，所以用暴力。对于每一个数都去判断它是否由对应的位置，有就把它放进去，并且标记这个位置已被占用。
但我感觉这种解法是错的，因为有可能是这种情况，假设在一串数中有两个数A B，A可以有两个位置（10st、15st）可以放，但B只能放到10st，根据我的解法，A会占用10st，B无处可放。
假AC代码：
#include <iostream>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<deque>
#include<stack>
using namespace std;
bool cmp(int a,int b)
{
return a<b;
}
int main()
{
int t,n,k,i,p,j,flag=0;
int con[110];
int check[110];
scanf("%d",&t);
for(i=1;i<=t;i++)
{
flag=0;
memset(check,0,sizeof(check));
scanf("%d%d",&n,&k);
for(j=1;j<=n;j++)
scanf("%d",&con[j]);
sort(con+1,con+n+1,cmp);
for(j=1;j<=n;j++)
{
while(1)
{
if(con[j]>n)
break;
if(check[con[j]]==0)
{
check[con[j]]=1;
flag++;
break;
}
con[j]+=k;
}
}
if(flag==n)
printf("Jerry\n");
else
printf("Tom\n");
}
return 0;
}
