C - XOR to all (binary topic)

https://atcoder.jp/contests/arc135/tasks/arc135_c
#include<iostream>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
const int N = 35;

int cnt[N];
int a[300005];
int n;
void solve() {
	cin >> n;ll ans = 0;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		int t= a[i];
		ans += t;
		for (int j  =0; j <= 30; j ++)
			cnt[j] += (t >> j & 1);
	}
	
	for (int i = 1; i <= n; i++) {
		ll res = 0;
		for (int j = 0; j <= 30;  j++)
			res += ((a[i] >> j & 1) == 1? n - cnt[j] : cnt[j]) * ((ll)1 << j);
		ans = max (ans, res);
	}
	
	cout << ans << endl;
}
int main () {
    int t;
    t =1;
    while (t --) solve();
    return 0;
}

The difficulty lies in how to efficiently count the sum after XOR . We observe binary , For each selected value , We have found , If the binary bit of the current value is 0, Then the binary of the same position of all other numbers is upper 1 Will be preserved , And contribution equals 1 Multiply the number of by the weight of the position . Otherwise, it counts 0 The number of contributions . Pay attention to the synthesis of the initial state . I've been thinking about how to make a sum , Numbers are not considered as binary bits , So we can't find the law . Next time you encounter this problem, you should consider binary representation .

 Some properties of XOR 
 For sequence A, B,B It's from A The result after XOR operation .
a[i]^a[j] = b[i]^b[j];