

如果n为奇数,k为奇数,LZC先手一定拿剩下k+1,LZC赢,k为偶数,LZC拿剩下k,LZC赢。
如果n为偶数,k为n-1,LZC先手只能拿1,HGL赢
#include<cstdio>
#include<iostream>
using namespace std;
int n,m;
int a[10010],b[10010];
int main()
{
cin>>n;
if(n%2)
{
cout<<"-1"<<endl;
}
else
cout<<n-1<<endl;
}
#include<cstdio>
#include<iostream>
using namespace std;
int n,m;
int a[10010],b[10010];
int main()
{
cin>>n;
if(n%2)
{
cout<<"-1"<<endl;
}
else
cout<<n-1<<endl;
}
#include<cstdio> #include<iostream> using namespace std; int n,m; int a[10010],b[10010]; int main() { cin>>n; if(n%2) { cout<<"-1"<<endl; } else cout<<n-1<<endl; }