HDU 4764-Stone(博弈)

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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

题意：

J 和 T 轮流说数字，后一个数字Y 与前一个数字 X 应该符合 1<=Y - X <=k ; 谁说的数字大于等于 N , 则输， J 先。

思路：

简单的博弈题。采用逆推的想法，首先数到 N 和大于 N 的数就输，因为两人都很聪明，则一个数到N-1，则另一个人就输了，则这个位置叫做必胜位。再往前推，如果有人走到了N-1-1 到 N-1-k 的位置，则另一个人肯定会在 N-1 的位置，则这个人就会输了。则这些个位置是必输点。

样例：N= 10，K= 2

0 1 2 3 4 5 6 7 8 9 10

Y N N Y N N Y N N Y N

可以发现0 位置就是J的比赛结果，必输必赢是周期为（k+1）的循环，所以可以取模。

CODE:

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;bool check(int n, int k){    return (n%(k+1) == 1);}int main(){    int N,K;    while(~scanf("%d %d", &N, &K),N){        if(check(N, K)) {            printf("Jiang\n");        }        else{            printf("Tang\n");        }    }    return 0;}

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