HDU 1846 Brave Game 博弈水题

来源：互联网发布：python运维开发视频编辑：程序博客网时间：2024/05/22 05:14

题目大意：

就是简单的取石子游戏....n个石子两个人每次可以去1~m个, 取走最后的胜利

大致思路:

如果n % (m + 1) == 0后手胜，否则先手胜...没什么好解释的了...

代码如下：

Result : Accepted Memory : 1620 KB Time : 0 ms

/* * Author: Gatevin * Created Time:  2015/4/28 21:06:00 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endlint T;int n, m;int main(){    scanf("%d", &T);    while(T--)    {        scanf("%d %d", &n, &m);        if(n % (m + 1) == 0) puts("second");        else puts("first");    }    return 0;}

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