I’ve got AC, but it’s too slow and my rating didn’t increase too much.

## solution

Do bitwisely. $O(N)$.

Let the given nonnegative numbers be $x_0, x_1, \dots, x_{k-1}$, and there are $n-k$ masked numbers. Let the actual numbers be $a_0, a_1, \dots, a_{k-1}, \dots, a_{n-1}$ and $X$ be the xor-sum of them, then there are $k$ equations: $\begin{array}{ll} a_i = X \oplus x_i & (i \lt k) \end{array}$. The definition of $X$ is: $X = \Sigma^{\oplus}\_i a_i$.

Assuming $a_k, \dots, a_{n-1}$ are $0$, the sum of $x_i$ makes $X$, $\begin{array}{ll} a_0 \oplus a_1 \oplus \dots \oplus a\_{k-1} = X = x_0 \oplus x_1 \oplus \dots \oplus x\_{k-1} & (k \; \text{is even}) \\\\ a_0 \oplus a_1 \oplus \dots \oplus a\_{k-1} = X = x_0 \oplus x_1 \oplus \dots \oplus x\_{k-1} \oplus X & (k \; \text{is odd}) \end{array}$. In the case that $k$ is even, the $X$ is fixed and you can compute the answer. In the case that $k$ is odd, if the $x_0 \oplus x_1 \oplus \dots \oplus x_{k-1}$ is $0$ then $X$ is arbitrary. Otherwise, contradiction. But also, if you have $x_k$, a number which has no constraint, you can use this to make $X$ arbitrary.

Let think about deciding the free $X$ (without $a_k$, and with $a_k$). When $a_k$ is not used, bitwisely do. Count $1$s of the $i$-th bit of $x_k$, and define the $i$-th bit of $X$ according to the result. When $a_k$ is used, do almost same thing, but you sholud take care about how to decide the $x_k$.

## implementation

#include <bits/stdc++.h>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define whole(f,x,...) ([&](decltype((x)) y) { return (f)(begin(y), end(y), ## __VA_ARGS__); })(x)
typedef long long ll;
template <class T> void setmin(T & a, T const & b) { if (b < a) a = b; }
using namespace std;
class OthersXor { public: long long minSum(vector<int>); };

const ll inf = ll(1e18)+9;
long long OthersXor::minSum(vector<int> xs) {
int n = xs.size();
xs.erase(whole(remove, xs, -1), xs.end());
auto foo = [&](ll sum) {
ll acc = 0;
for (int x : xs) acc += sum ^ x;
return acc;
};
auto bar = [&]() {
vector<int> cnt(32);
for (int x : xs) {
repeat (i, cnt.size()) {
cnt[i] += (x & (1ll<<i)) != 0;
}
}
ll acc = 0;
repeat (i, cnt.size()) {
if (cnt[i] > xs.size()/2) {
acc |= 1ll<<i;
}
}
return acc;
};
ll ans = inf;
ll sum_x = whole(accumulate, xs, 0ll, bit_xor<ll>());
if (xs.size() % 2 == 0) {
setmin(ans, foo(sum_x));
if (xs.size() < n) {
xs.push_back(sum_x);
setmin(ans, foo(bar()));
xs.pop_back();
}
} else {
if (sum_x == 0) {
setmin(ans, foo(bar()));
}
if (xs.size() < n) {
ll sum = bar();
xs.push_back(sum ^ sum_x);
setmin(ans, foo(sum));
xs.pop_back();
}
}
if (ans == inf) ans = -1;
return ans;
}