hdu 4764 Stone

Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1539    Accepted Submission(s): 1100

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

1 130 310 20 0

 

Sample Output

JiangTangJiang

题目大意：Tang和Jiang轮流写数字，Tang先写；如若写的数x满足1<=x<=k，则下次写的数y满足1<=y-x<=k，谁先写到不小于n的数算输。

思路：hpu王小二点击打开链接（第二种）此题为谁先越界谁输，把此题具体化为取物品，即为谁先取完物品谁输；

x+1<=y<=k+x，取物品的问题，谁最后取完谁输，每次最少取一个，最多取k个；

#include<cstdio>int main(){int n,k;while(scanf("%d %d",&n,&k) && (n+k)){if((n-1)%(k+1)==0) printf("Jiang\n");else printf("Tang\n");}return 0;}