Game with Pearls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 414 Accepted Submission(s): 212

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）

题目大意：Jerry 和 Tom 玩一个游戏，给你 n 个盒子， a[ i ] 表示开始时，第 i 个盒子中的小球的个数。然后 Jerry 可以在每个

盒子里加入 0 或 k的倍数的小球，操作完后，Jerry 可以重新排列盒子的顺序，最终使第 i 个盒子中有 i 个小球。若Jerry能

使最终的盒子变成那样，就输出 “Jerry” ，否则输出 “Tom” 。

解题思路：

方法一、统计后，从小到大分过去，剩下的放到i+k的位置，这样O（n）的可以得到答案。

方法二、转化乘二部图模型，将初始状态和可到达的球数作为两个集合，符合倍数条件的连边，若最大匹配数=n，则Jerry赢。

#include 
  
   
#include 
   
     using namespace std; const int MAXN = 110; int nCase, n, k, m, cnt[MAXN]; void init() { memset(cnt, 0, sizeof(cnt)); } void input() { cin >> n >> k; for (int i = 0; i < n; i++) { int x; cin >> x; cnt[x]++; } } void solve() { for (int i = 1; i <= n; i++) { if (!cnt[i]) { cout << "Tom" << endl; return; } cnt[i+k] += cnt[i] - 1; } cout << "Jerry" << endl; } int main() { ios::sync_with_stdio(false); cin >> nCase; while (nCase--) { init(); input(); solve(); } return 0; }

HDU 5090 Game with Pearls（贪心, 二部图最大匹配）

Game with Pearls