博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
hdoj--2509--Be the Winner(尼姆博弈)
阅读量:4615 次
发布时间:2019-06-09

本文共 1562 字,大约阅读时间需要 5 分钟。

Be the Winner

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2718    Accepted Submission(s): 1487
Problem Description
Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).
 
Input
You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.
 
Output
If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
 
Sample Input
 
2 2 2 1 3
 
Sample Output
 
No Yes
#include 
#include
using namespace std;#define M 50int num[M];int main(){ int i,ans,re,n; while(scanf("%d",&n)!=EOF) { ans=0,re=0; for(i=0;i
1) ans++; re^=num[i]; } if(ans) { if(re) printf("Yes\n"); else printf("No\n"); } else { if(re) printf("No\n"); else printf("Yes\n"); } } return 0;}

转载于:https://www.cnblogs.com/playboy307/p/5273783.html

你可能感兴趣的文章
css3转换
查看>>
读取文件,按行输出
查看>>
向PHP发送HTTP-Get请求
查看>>
本周作业
查看>>
rsyslog 传输日志
查看>>
apache性能优化
查看>>
UNDO 100%
查看>>
MaReduce理论
查看>>
CentOS7安装RabbitMQ
查看>>
mysql乐观锁和悲观锁
查看>>
将字符串中不同字符的个数打印出来
查看>>
java第三次上机
查看>>
android Javah生成JNI头文件
查看>>
npm创建react项目
查看>>
关于u32中查找和定位最后到bit Number of 1 Bits
查看>>
sql数据库查询
查看>>
云计算技能图谱
查看>>
js的块级作用域
查看>>
委托、Lambda表达式和事件
查看>>
typecho模板制作代码收集
查看>>