hdu 4763 stone(博弈)

F - Stone

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

 

 

简单博弈 运用逆推

要赢得胜利就必须占据 n-1 点 而且要避开 n-1-k ~n-2 之间的点

以此类推 可以推出 只要 （n-1）%（k+1） 不为0  第一个人就能胜利

 

#include<cstdio>

int main()

{

    int n,k;

    while(scanf("%d%d",&n,&k)!=EOF)

    {

        if(n==0&&k==0) break;

        int que=(n-1)%(k+1);

        if(que>0&&que<=k) printf("Tang\n");

            else printf("Jiang\n");

    }

    return 0;

}

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