【JZOJ4909】李电下棋

Description

Solution

这题我们可以找规律，是gcd(a,b)倍数的格子都可以被下到。

证明： gcd(a,b)=gcd(a,a−b)=gcd(a,a+b)

Code

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define ll long long
using namespace std;
ll gcd(ll x,ll y)
{
    ll z;
    while(x%y!=0)
    {
        z=x%y;
        x=y;
        y=z;
    }
    return y;
}
int main()
{
    freopen("chess.in","r",stdin);
    freopen("chess.out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n,a,b;
        scanf("%lld %lld %lld",&n,&a,&b);
        if(n/gcd(a,b)%2==1) printf("lidian");
        else printf("wfl");
        printf("\n");
    }
}