HDU 5090 Game with Pearls

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1196    Accepted Submission(s): 441

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 

 

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5090

（以下题意来自网上）

题目大意：

　　　　Tom和Jerry做游戏，给定N个管子，每个管子上面有一定数目的珍珠。

　　　　现在Jerry开始在管子上面再放一些珍珠，放上的珍珠数必须是K的倍数，可以不放。

　　　　最后将管子排序，如果可以做到第i个管子上面有i个珍珠，则Jerry胜出，反之Tom胜出。

解题思路：

　　　　贪心：（15ms）

　　　　　　　将N个管子按管子中的珍珠数量升序排序，则满足条件的时候第i个管子应该有i个珍珠。

　　　　　　　每个管子开始遍历，如果第i个管子有i个珍珠，则进行下一次循环；

　　　　　　　如果小于i个珍珠，则加上k个珍珠，重新按升序排序；//------------------------注意这里要重新排序--------所以我之前一直在WA着----

　　　　　　　如果大于i个珍珠，则说明不满足条件 .

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int num[1000];int main(){int t;cin >> t;while(t--){int n,k;cin >> n >> k;for(int i=1;i<=n;i++)scanf("%d",&num[i]);int tot = 0;int flag = 1;sort(num+1,num+(n+1));for(int i=1;i<=n;i++){while(num[i]!=i&&num[i]<i){num[i]+=k;sort(num+i,num+(n+1));}if(num[i]==i)continue;else if(num[i]>i){flag = -1;break;}}if(flag==1)cout<<"Jerry\n";else cout<<"Tom\n";}return 0;}