Petrozavodsk Winter Training Camp 2018 Jagiellonian U Contest Problem A. XOR

First of all the number of exclusive-OR sum and sum up to give some number of bits is 1 bit 0 is a 0 bit is a number of all there is an even number the number of

No matter how you divide the number of this one on the final answers are not contributing because of the even = even + even / odd + odd

So we put all direct & sum regardless of the number of those bits do not contribute to the re-insertion as a linear base

Come up with a maximum of 1 bit to A then we have to do now is to try to ensure that A high can not rest on the other 1 is 1 so the answer is optimal

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[65];
ll num[100005];
ll x;
int main () {
         int TNT;
        scanf("%d", &TNT);
        while (TNT--) {
                memset(a, 0, sizeof(a));
                int n;
                ll sum = 0;
                scanf("%d", &n);
                for (int i = 1; i <= n; i++) {
                        scanf("%lld", &x);
                        Surely, [i] = x;
                        sum ^= x;
                }
                for (int i = 1; i <= n; i++) {
                        num[i] &= sum;
                }
                for (int i = 1; i <= n; i++) {
                        for (int j = 62; j >= 0; j--) {
                                if (num[i] & (1ll << j)) {
                                        if (!a[j]) {
                                                a[j] = num[i];
                                                break;
                                        } else {
                                                num [i] '= a [j];
                                        }
                                }
                        }
                }
                int aim = -1;
                ll aimx = 0;
                for (int i = 62; i >= 0; i--) {
                        if (sum & (1ll << i)) {
                                aim = i;
                                aimx = a[i];
                                break;
                        }
                }
                for (int i = aim - 1; i >= 0; i--) {
                        if (aimx & (1ll << i)) {
                                aimx ^= a[i];
                        }
                }
                ll ans = aimx - (aimx ^ sum);
                printf("%lld\n", ans);
        }
}