题意:

  给你n个数字(<230),求出一个数字使得所有数字^这个数字之后,二进制状态下的1的个数相同。

解析:

  因为最大数字二进制数有30位,所以分为前15位和后15位,分别枚举0-1<<15,用map匹配。

代码:

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <set> using namespace std; typedef long long ll; const int maxn=+; ll a[maxn];
map<vector<int>, ll> mp; int main(){
ll n;
scanf("%lld",&n); for(int i=;i<n;i++)scanf("%lld",&a[i]);
for(int i=;i< (<<);i++){
vector<int> vd(n);
for(int j=;j<n;j++)vd[j]= __builtin_popcount( (a[j] & ) ^i)- __builtin_popcount((a[]&)^i);
if(not mp.count(vd))mp[vd]=i;
}
for(int i=;i< (<<);i++){
vector<int> vd(n);
for(int j=;j<n;j++)vd[j]= __builtin_popcount( (a[]>>) ^i)- __builtin_popcount( (a[j]>>) ^i);
if(mp.count(vd)){ printf("%lld\n",(i<<) + mp[vd]);
return ;
}
}
printf("-1\n");
return ;
}

  

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