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
题意：给出n和k
t和j每次给出一个数 相差不超过k 并且递增
t先进行 并且第一次范围 在 1到k之间
先到达n的人输
思路:
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
以 10 2 为例
1 2 3 4 5 6 7 8 9 10
l l w l l w l l w
#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 "); else printf("Jiang "); } return 0; }