Codeforces Round #425 A
题意:n根棍子,每次拿走k根,当不足k根的时候不能拿,输,判断先手能否赢
思路:(n/k)&1,直接判断能经行奇数次取木棍还是偶数
AC代码:
#include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #define ll long long #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a) memset(a,0,sizeof(a)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define lrt (root*2) #define rrt (root*2+1) #define len (r-l+1) #pragma comment(linker, "/STACK:102400000,102400000") using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; ll n,k; int main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>n>>k; ll ans=n/k; if(ans&1) cout<<"YES"; else cout<<"NO"; return 0; }