POJ2975 Nim

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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 npiles having k₁, k₂, …, k_n stones respectively; in such a position, there are k₁ + k₂ + … + k_n possible moves. We write each k_i in binary (base 2). Then, the Nim position is losing if and only if, among all the k_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 the xor 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:

 111
1011
1101

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 ≤ k_i ≤ 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

3
7 11 13
2
1000000000 1000000000
0

Sample Output

3
0

Source

Stanford Local 2005

 

 

这次是经典的nim问题，求先手若要必胜，有多少堆石头可以取。

若先手可以取走一些石头，使得剩下每堆的石头数异或和为0，那么算一种可行的方法。

首先若是先手可以必胜，那么所有堆石头数异或和s==0。若要从第i堆中取，使之后异或和为0，那么需要取s^a[i]堆((s^a[i])^a[i]==s==0)

若石头数量a[i]>(s^a[i])，就可以取。

 

另外要注意异或优先级比小于号低，需要加括号

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int n;
int a[1200],s,ans;
int main(){
    while(scanf("%d",&n) && n){
        int i,j;
        s=0;ans=0;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]),s^=a[i];
        for(i=1;i<=n;i++){
            if((s^a[i])<a[i]) ans++;
        }    
        printf("%d
",ans);
    }
    return 0;
}