hdu4764 Stone 巴什博弈

Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1439    Accepted Submission(s): 1014

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

 

最后肯定剩一个给对方，谁拿谁输，对N - 1进行巴什博弈，谁先拿完前N - 1也就是剩一个给对方，谁就赢

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int n, k;

int main()
{
	while (~scanf("%d%d", &n, &k) && (n || k)) {
		n -= 1;
		if (n % (k + 1) != 0) {
			puts("Tang");
		} else {
			puts("Jiang");
		}
	}
	return 0;
}