Codeforces Round #613 (Div. 2) D. Dr. Evil Underscores

贪心 + 分治 + 位运算
从高到低逐位考虑，分两种情况：
1、目前集合中当前位只有0或只有1，此位贡献为0
2、目前集合中当前位(第k位)同时存在0和1，那么此位对这个集合的贡献为 1 << k，然后以0和1划分当前集合的子集，分别求子集的最小贡献

#include<bits/stdc++.h>
using namespace std;

#define fr first
#define se second

typedef pair<int, int> PII;
typedef long long LL;
typedef unsigned long long ULL;

const int INF = 0X3f3f3f3f, N = 1e5 + 10, MOD = 1e9 + 7;

vector<int> v;

int get_min(int k, vector<int> & cur) {
	if (k < 0) return 0;
	vector<int> a, b;
	for (int i = 0; i < cur.size(); i++) {
		if (cur[i] >> k & 1) a.push_back(cur[i]);
		else b.push_back(cur[i]);
	}
	if (a.size() == 0) return get_min(k - 1, b);
	if (b.size() == 0) return get_min(k - 1, a);
	return min(get_min(k - 1, a) + (1 << k), get_min(k - 1, b) + (1 << k));
}

void work() {
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		int x;
		cin >> x;
		v.push_back(x);
	}

	cout << get_min(31, v);
}

signed main() {

	work();

	return 0;
}

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来源： https://www.cnblogs.com/xhy666/p/16425466.html