Brave Game(巴什博奕)

发布于 2020-10-29  0 次阅读


题目大意:

Brave Game
有一堆石子一共有n个;
两人轮流进行;
每走一步可以取走1…m个石子;
最先取光石子的一方为胜;

思路:

很明显的巴什博弈
n对m+1取模判断余数是否为0即可

AC Code:

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<math.h>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
// #define int long long
#define debug(a) cout<<#a<<"="<<a<<endl;
typedef long long ll;
const double PI=acos(-1.0);
const double e=exp(1.0);
const int M=1e9+7;
const int N=2e5+7;
inline int mymax(int x,int y){return x>y?x:y;}
inline int mymin(int x,int y){return x<y?x:y;}

void solve(){
    int n, m;
    cin>>n>>m;
    if(n%(m+1)==0)  cout<<"second"<<endl;
    else            cout<<"first"<<endl;
	return ;
}

signed main(){
    int T;cin>>T;
    while(T--) solve();
    return 0;
}

平平无奇的大学在读本科生