cf 613.D（分治+暴力）

最新推荐文章于 2022-02-13 20:31:55 发布

原创最新推荐文章于 2022-02-13 20:31:55 发布 · 368 阅读

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本文介绍了一种算法，用于解决给定一组整数，如何选择一个整数X，使得所有整数与X进行异或操作后的最大值最小化的问题。通过使用位运算和分治策略，文章详细解释了实现这一目标的高效算法，并提供了完整的C++代码示例。

Today, as a friendship gift, Bakry gave Badawy n integers a1,a2,…,an and challenged him to choose an integer X such that the value max_1≤i≤n(ai⊕X).is minimum possible, where ⊕ denotes the bitwise XOR operation.

As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of max_1≤i≤n(ai⊕X).

Input
The first line contains integer n (1≤n≤10⁵).

The second line contains n integers a1,a2,…,an (0≤ai≤2³⁰⁻¹).

Output
Print one integer — the minimum possible value of max_1≤i≤n(ai⊕X).

Examples
input
3
1 2 3
output
2

input
2
1 5
output
4

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) ((x)<<1)
#define r(x) ((x)<<1|1)
#define lowbit(x) ((x)&(-(x)))
#define abs(x) ((x)>=0?(x):(-(x)))
#define ms(a,b) memset(a,b,sizeof(a))
#define NSYNC std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);

using namespace std;

int n, a[111111];

int get(int low, int hei, int bit) {
	if (bit < 0)return 0;
	int mask = 1 << bit;
	int f = -1;
	for (int i = low; i < hei; i++) {
		if (a[i] & mask) {//找到第bit位为1的，就给f赋值
			f = i;
			break;
		}
	}

	if (f == -1) return get(low, hei, bit - 1);
	//第一种情况，到了hei都没有找到bit位为1的，所以所有的bit为都是0，所以直接计算bit-1
	if (f == low) return get(low, hei, bit - 1);
	//第二种情况，第一个就是bit位为1，那么后面的bit为都为1，直接计算bit-1，此时f=low

	return mask | min(get(low, f, bit - 1), get(f, hei, bit));
}

int main() {
	NSYNC;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n);
	cout << get(0, n, 30);
	return 0;
}