51nod1066bash游戏

题目：一堆石子有n个， 每次可以取大于等于1不超过m的任意个，去最后一个的人胜利，给你n , m ，问先手必胜还是后手必胜。

思路：最优解是两个人每回合取（m+1）个数。所以只需要判断n%（m+1)的结果就可以

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <fstream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <iomanip>using namespace std;//#pragma comment(linker, "/STACK:102400000,102400000")#define maxn 200005#define MOD 1000000007#define mem(a , b) memset(a , b , sizeof(a))#define LL long long#define ULL long longconst long long INF=0x3fffffff;int main(){    int t  , n , m;    cin >> t;    while(t--)    {        scanf("%d %d" , &n , &m);        if(n % (m + 1)) cout << "A" << endl;        else cout << "B" << endl;    }    return 0;}