POJ 题目2975 Nim(nim)
来源:互联网 发布:九次方大数据如何使用 编辑:程序博客网 时间:2024/05/29 08:23
Description
Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position.
A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains n piles having k1, k2, …, kn stones respectively; in such a position, there are k1 + k2 + … + kn possible moves. We write each ki in binary (base 2). Then, the Nim position is losing if and only if, among all the ki’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the xor of the ki’s is 0.
Consider the position with three piles given by k1 = 7, k2 = 11, and k3 = 13. In binary, these values are as follows:
11110111101
There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, suppose k3 were changed to be 12. Then, there would be exactly two 1’s in each digit position, and thus, the Nim position would become losing. Since a winning move is any move that leaves the game in a losing position, it follows that removing one stone from the third pile is a winning move when k1 = 7, k2 = 11, and k3 = 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles.
Input
The input test file will contain multiple test cases, each of which begins with a line indicating the number of piles, 1 ≤ n ≤ 1000. On the next line, there are n positive integers, 1 ≤ ki ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case with n = 0 and should not be processed.
Output
For each test case, write a single line with an integer indicating the number of winning moves from the given Nim position.
Sample Input
37 11 1321000000000 10000000000
Sample Output
30
Source
Stanford Local 2005
题目比较水,读题读好久
题目大意就是:n堆石子,每堆可以去任意个,问先手赢时,取第一步的取法有多少种
ac代码
#include<stdio.h>int main(){int n;while(scanf("%d",&n)!=EOF,n){int a[1010];int i,sum=0,ans=0;for(i=0;i<n;i++){scanf("%d",&a[i]);sum^=a[i];}ans=0;for(i=0;i<n;i++){if(a[i]>(sum^a[i])){ans++;}}printf("%d\n",ans);}}
- POJ 题目2975 Nim(nim)
- POJ 2975 Nim(nim博弈)
- POJ-2975-Nim(Nim博弈)
- [POJ](2975)Nim ---Nim博弈(博弈)
- POJ 2975 Nim <Nim 博弈>
- POJ 2975 Nim(博弈论)
- POJ 2975 Nim(博弈)
- 经典的NIM-poj-2975-Nim
- POJ 2975 Nim (Nim的证明)
- POJ-2975-Nim
- poj 2975 Nim 水
- POJ 2975 Nim
- POJ Nim (2975)
- poj 2975 Nim
- POJ 2975 Nim题解
- 【POJ 2975】 Nim 博弈论
- poj 2975 Nim
- poj 2975 Nim(博弈)
- 在emacs里面安装js2-mode
- 黑马程序员——Java(反射)
- HDU 1896 -- Stones (优先队列)
- Android程序员必备精品资源
- 阅读程序写出运行结果7
- POJ 题目2975 Nim(nim)
- SQLSERVER 数据库引擎中的锁定
- 网络编程(获取json和网络图片)
- fakenet
- SQLSERVER 锁粒度和层次结构
- 阅读程序写出运行结果8
- 网络编程(自定义缓存图片)
- SQLSERVER锁模式
- JAVA菜鸟入门篇 - 封装/隐藏、java访问控制符 (十七)