一共n个石头,两人轮流取1~m个石头,问先手必胜还是后手必胜
设n = (m + 1) * r + s;(0<= s <= m)
若A取了石头过后,s == 0,则A必胜
#include <cstdio>
using namespace std;
int main()
{
int t, n, m;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &m);
if(n % (m + 1) == 0) printf("second\n");
else printf("first\n");
}
return 0;
}