D1. 388535 (Easy Version)

// Problem: C. Shinju and the Lost Permutation
// Contest: Codeforces - Codeforces Round #779 (Div. 2)
// URL: https://codeforces.com/contest/1658/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-27 22:59:43
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
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 mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a;
		a = a * a;
		b >>= 1;
	}
	return ans;
}
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');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
void solve() {
	int l, r;
	cin >> l>> r;
	int cnt[32] = {0};
	for (int i = l; i <= r; i ++) {
		int x;
		cin >> x;
		bitset<32> c(x);
		for (int j = 0; j < 32; j ++)
			cnt[j] += c[j]; 
	}
	int ans=  0;
	for (int i= 0; i < 32; i ++) {
		if (cnt[i] > r - l + 1 - cnt[i])
			ans += qmi(2, i);
	}
	print(ans);
	puts("");
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

通过研究和法的排列可以发现针对每一位，如果cnt[i] * 2 > r - l+1,说明有多余的1，那么就需要要用1来异或掉。由此可以得到做法