Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 179    Accepted Submission(s): 137

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  

Source
2013 ACM/ICPC Asia Regional Changchun Online  

Recommend liuyiding  

很水的博弈，和取石子游戏是一样的。

必败点是 (n-1)%(k+1)==0

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/9/28 星期六 12:09:01
 4 File Name     :2013长春网络赛\1006.cpp
 5 ************************************************ */
 6 
 7 #pragma comment(linker, "/STACK:1024000000,1024000000")
 8 #include <stdio.h>
 9 #include <string.h>
10 #include <iostream>
11 #include <algorithm>
12 #include <vector>
13 #include <queue>
14 #include <set>
15 #include <map>
16 #include <string>
17 #include <math.h>
18 #include <stdlib.h>
19 #include <time.h>
20 using namespace std;
21 
22 int n,k;
23 int main()
24 {
25     //freopen("in.txt","r",stdin);
26     //freopen("out.txt","w",stdout);
27     while(scanf("%d%d",&n,&k) == 2)
28     {
29         if(n == 0 && k == 0)break;
30         if((n-1)%(k+1) == 0)printf("Jiang\n");
31         else printf("Tang\n");
32     }
33     return 0;
34 }