HDU 4764-Stone(博弈)

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each case, print the winner's name in a single line.

 

[align=left]Sample Input[/align]

1 1
30 3
10 2
0 0

 

[align=left]Sample Output[/align]

Jiang
Tang
Jiang

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;
}