【HDOJ 4764】 Stone (博弈)

【HDOJ 4764】 Stone (博弈)

Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1090    Accepted Submission(s): 761

Problem Description

Tang and Jiang are good friends. To decide whose treat it is for dinner, they are playing a game. Specifically, Tang and Jiang will alternatively write numbers (integers) on a white board. Tang writes first, then Jiang, then again Tang, etc... Moreover, assuming that the number written in the previous round is X, the next person who plays should write a number Y such that 1 <= Y - X <= k. The person who writes a number no smaller than N first will lose the game. Note that in the first round, Tang can write a number only within range [1, k] (both inclusive). You can assume that Tang and Jiang will always be playing optimally, as they are both very smart students.

 

Input

There are multiple test cases. For each test case, there will be one line of input having two integers N (0 < N <= 10^8) and k (0 < k <= 100). Input terminates when both N and k are zero.

 

Output

For each case, print the winner's name in a single line.

 

Sample Input

   
   
   
   

    
    
    
    1 1
30 3
10 2
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Jiang
Tang
Jiang
   
   
   
   

 

写数字游戏 轮流写出小于n的一个数 数字要求递增 并且当前玩家写出的数与前一玩家写的数差值 <= k  写出n的输

可以转换为取石子(然而题目就是Stone 为了不太赤果果 给改编了下= =也醉了

因此 只要(n-1)为(k+1)的倍数 即后手始终可以达到 k+1 2*(k+1) 3*(k+1)....的状态 这样先手必输 其余状态先手可达n-1

代码如下:

#include <bits/stdc++.h>

using namespace std;

int main()
{

	int n,m,t,i,j,u,v,w;
	while(~scanf("%d %d",&n,&m) && n+m)
	{
		if((n-1)%(m+1)) puts("Tang");
		else puts("Jiang");
	}


    return 0;
}