题目意思

给出的个数，给出次询问，每次询问都会给出两个左右区间，问我们为什么时候异或区间全部数之后答案最大。

解题思路

异或基本思想，
所以统计这个区间中这一位共有几个1，如果大于0的个数那么这一位就取0，否则0个数大于1个数就取1在的这一位。
有关区间各个位数统计可以用前缀和预处理，把时间复杂度降为

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 1e9 + 7;
const ll INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
ll bit[N][35];

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i) {
        int a = read();
        for (int j = 0; j < 31; ++j) { // 1~n全部数第j位中1的个数
            if (a & 1)    bit[i][j] = bit[i - 1][j] + 1;
            else bit[i][j] = bit[i - 1][j];
            a >>= 1;
        }
    }
    int m = read();
    while (m--) {
        int ans = 0;
        int l = read(), r = read(), len = r - l + 1;
        for (int i = 0; i < 31; ++i) {
            int k = bit[r][i] - bit[l - 1][i];
            if (k < len - k)    ans |= 1 << i; //如果1个数大于0个数就取1
        }
        write(ans), putchar(10);
    }
    return 0;
}