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POJ2975 Nim（尼姆博弈）
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Armin's Blog
POJ2975 Nim（尼姆博弈）
March 09, 2016
题意：n 个数分别代表每堆的石子数，问获胜的取法有多少种。
和HDU1850一样的代码。。
简单的再总结下，就是用异或的和 sum 先异或 ai 这堆，由于 a^b^b=a，那么就相当于没考虑这一堆，所以只要把 ai 这堆剩下 sum^ai 个石子，那么所有堆就变成了(sum^ai)^(sum^ai)=0，对手就面临着必败态。所以让 sum^ai 小于 ai，ans++就可以。也就是说每堆最多只有一种取法。
#include<cstring>
#include<cstdio>
#include<cstring>
#include<cmath>
#include <vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
int a[1005];
int main(){
//freopen("a.txt", "r", stdin);
int n;
while(cin >> n && n){
int sum = 0;
for(int i = 0; i < n; i++){
scanf("%d", &a[i]);
sum ^= a[i];
}
int ans = 0;
for(int i = 0; i < n; i++){
if((sum^a[i]) < a[i])
ans++;
}
cout << ans << endl;
}
return 0;
}
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