D. Dr. Evil Underscores

最新推荐文章于 2021-04-01 19:48:49 发布

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本文解析了CodeForces竞赛中一道关于寻找最小最大异或值的问题，通过深入分析给出了一种高效的解决策略，利用二分查找和动态规划思想，实现了对给定整数数组的最优解求解。

链接：https://codeforces.com/contest/1285/problem/D

Today, as a friendship gift, Bakry gave Badawy nn integers a1,a2,…,ana1,a2,…,an and challenged him to choose an integer XX such that the value max1≤i≤n(ai⊕X)max1≤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 max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).

Input

The first line contains integer nn (1≤n≤1051≤n≤105).

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤230−10≤ai≤230−1).

Output

Print one integer — the minimum possible value of max1≤i≤n(ai⊕X)max1≤i≤n(ai⊕X).

Examples

input

Copy

3
1 2 3

output

Copy

input

Copy

2
1 5

output

Copy

Note

In the first sample, we can choose X=3X=3.

In the second sample, we can choose X=5X=5.

代码：

#include <bits/stdc++.h>
using namespace std;
long long n;
long long a[100010];
long long dfs(long long p, long long q, long long h) 
{
	if(h==-1) 
	return 0;
	long long l=p,r=q,t=p-1;
	while(l<=r) 
	{
		long long mid=(l+r)/2;
		if((a[mid]&((1<<h+1)-1))<(1<<h)) 
		l=(t=mid)+1;
		else 
		r=mid-1;
	}
	if(t==p-1||t==q) 
	{
		return dfs(p,q,h-1);
	}
	return min(dfs(p,t,h-1),dfs(t+1,q,h-1))+(1<<h);
}
int main() 
{
	scanf("%lld",&n); 
	for(int i=1;i<=n;i++) 
	{
		scanf("%lld",&a[i]);
	}
	sort(a+1,a+n+1);
	printf("%lld\n",dfs(1,n,30));
}