HDU Rabbit and Grass 兔子和草 (Nim博弈)
2015-05-12 11:55
423 查看
思路:简单Nim博弈,只需要将所给的数字全部进行异或,结果为0,则先手必败。否则必胜。
AC代码
#include <iostream> using namespace std; int main() { //freopen("input.txt", "r", stdin); int t, n, m; while(cin>>t,t) { int ans,tmp; cin>>ans; for(int i=1; i<t; i++) { cin>>tmp; ans^=tmp; } if(ans) cout<<"Rabbit Win!"<<endl; else cout<<"Grass Win!"<<endl; } return 0; }
AC代码
相关文章推荐
- HDU - 1849 Rabbit and Grass Nim 博弈入门题
- HDU 1849 Rabbit and Grass 【Nim博弈】
- hdu 1849 Rabbit and Grass(nim 博弈)
- hdu 1849 Rabbit and Grass(s-nim博弈)
- HDU 1849 Rabbit and Grass(nim博弈)
- HDU-1849-Rabbit and Grass【Nim博弈】
- HDU 1849 Rabbit and Grass【尼姆博弈】
- hdu 1849 Rabbit and Grass(nim游戏)
- Rabbit and Grass (nim game 尼姆 博弈)
- Rabbit and Grass + 博弈 + nim博弈
- HDU(1849)Rabbit and Grass(博弈)
- 杭电hdu 1849 Rabbit and Grass nim game
- hdu 1849 Rabbit and Grass SG定理(简单博弈)
- hdu 1849(Rabbit and Grass) 尼姆博弈
- hdu 1849 Rabbit and Grass(nim)
- HDOJ 题目1849Rabbit and Grass(nim博弈)
- HDU 1849 Rabbit and Grass(Nim game)
- HDU 1849 Rabbit and Grass 博弈Nim游戏
- hdu 1849 Rabbit and Grass
- HDU 1849 Rabbit and Grass(博弈入门)