Poj2975（NIM博弈）



Nim

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5380 Accepted: 2521

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 removingone 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, …, k_n stones respectively; in such a position, there arek₁ + k₂ + … +k_n possible moves. We write eachk_i in binary (base 2). Then, the Nim position is losing if and only if, among all thek_i’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 thexor of the k_i’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, supposek₃ 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 whenk₁ = 7, k₂ = 11, andk₃ = 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 ≤ k_i ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case withn = 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

题意：n堆物品，每个为ki个，问先手第一步有多少种做法使他获胜。

题解:

NIM的简单变形，先求出异或和ans，对于每一个ki，ans^ki为除i堆之外的和，假设在i堆取完物品后，i堆剩bi个，如果保证先手获胜，则ans^ki^bi=0.变形得ans^ki=bi,所以，在ki取ki-ans^ki个物品时在i堆能取胜的唯一方法，逐个判断，如果刚开始ans就等于0.则先手必输，输出0。

代码：

#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <queue>#include <iostream>using namespace std;int k[1005];int main(){    int t;    while(cin>>t)    {        if(t<=0)            break;        int ans=0;        for(int i=0;i<t;i++)        {            cin>>k[i];            ans^=k[i];        }        int sum=0;        for(int i=0;i<t;i++)        {            if((ans^k[i])<=k[i])                sum++;        }        if(ans==0)            sum=0;        cout<<sum<<endl;    }}