Arc135 C (the proof is not very clear)

subject
The question : Given array a, You can perform any operation . Select any number in the array a[i], Make the whole array execute ^a[i]. After the operation, the array a The largest sum of .
Ideas : It can be proved that the operation is not performed and the operation is performed 1 Times include all cases .

Time complexity : O(nlogn)
Code :

// Problem: C - XOR to All
// Contest: AtCoder - AtCoder Regular Contest 135
// URL: https://atcoder.jp/contests/arc135/tasks/arc135_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round()  rounding  ceil()  Rounding up  floor()  Rounding down 
// lower_bound(a.begin(),a.end(),tmp,greater<ll>())  First less than or equal to 
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 3e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){
     return x&(-x);}
template<typename T>void write(T x)
{
    
    if(x<0)
    {
    
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
    
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){
    if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){
    x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
int a[N];
int cnt[N][2]; //0、1
ll ans = 0;
void solve()
{
    
   read(n);
   for(int i=1;i<=n;++i)
   {
    
   	   read(a[i]);
   	   ans += a[i];
   	   for(int j=0;j<30;++j)
   	   {
    
   	   	  if(a[i]>>j&1) cnt[j][1]++;
   	   	  else cnt[j][0]++;
   	   }
   }
   for(int i=1;i<=n;++i)
   {
    
   	  ll res = 0;
   	  for(int j=0;j<30;++j)
   	  {
    
   	  	 if(a[i]>>j&1) res += 1ll*cnt[j][0]*(1<<j);
   	  	 else res += 1ll*cnt[j][1]*(1<<j);
   	  }
   	  ans = max(ans,res);
   }
   write(ans);
}
signed main(void)
{
      
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
    
   	 solve();
   }
   return 0;
}