“蔚来杯“2022牛客暑期多校训练营4 N

N

将答案的公式结合起来化简可得

         ,

这样不会爆long long，只需要处理出最终趋于稳定的数组的形式，可知a+b=a&b+a|b，操作无数次，总和是不变的

AC代码：

#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
using namespace std;
using LL = long long;

LL gcd(LL x, LL y) {
    return y == 0 ? x : gcd(y, x % y);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    LL n;
    cin >> n;
    vector<LL> a(n + 1);
    LL maxx = 0;
    LL sum = 0;
    vector<LL> pos(16, 0);
    for (int i = 1; i <= n; i ++ ) {
        cin >> a[i];
        sum += a[i];
    } 
    if (sum == 0) {
        cout << "0/1\n";
        return 0;
    }
    for (int i = 1; i <= n; i ++ ) {
        for (int j = 0; j <= 15; j ++ ) {
            if(1 << j & a[i]) {
                pos[j]++;
            }
        }
    }
    vector<LL> x;
    for (int i = 1; i <= n; i ++ ) {
        LL t = 0;
        for (int j = 15; j >= 0; j -- ) {
            if(pos[j]) {
                t += (1 << j);
                pos[j]--;
            }
        }
        if(t)
        x.push_back(t);
        else break;       
    }
    int len = x.size();
    LL ans = -sum * sum;
    for (int i = 0; i < n; i++) {
        if (i < len) 
            ans += n * x[i] * x[i];
        else {
            break;
        }
    }
    LL c = gcd(ans, n * n);
    cout << ans / c << "/" << n * n / c << '\n';

    return 0;
}