链接：http://codeforces.com/contest/1285/problem/D
D. Dr. Evil Underscores
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 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).
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 max1≤i≤n(ai⊕X).
Examples
inputCopy
3
1 2 3
outputCopy
2
inputCopy
2
1 5
outputCopy
4
Note
In the first sample, we can choose X=3.
In the second sample, we can choose X=5.
思路：
数据范围是2的30次方，我们在二进制下进行操作，最多有30位，所以我们直接从第30位进行考虑，对每一位来讲
分两种情况看：
- 这个二进制位上全是0或者全是1，这个时候当然 X 的二进制位也要是0或者1，保持相同异或才能是0嘛； 这个时候整个集合不变，去看下一位；
- 当这一位又有0又有1的时候，就把是0的分成一组，是1的分成一组，此时不管怎样异或最大值的二进制在这一位是1，然后再分别对这两个集合进行分情况讨论，也就是递归下去；
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector <int> a;
int get_ans(vector <int> &now,int bit)
{
if(now.size()==0||bit<0) return 0;
vector<int> zero,one;
for(int i=0;i<now.size();i++)
{
if((now[i]>>bit)&1) one.push_back(now[i]);
else zero.push_back(now[i]);
}
if(!one.size()) return get_ans(zero,bit-1);
if(!zero.size()) return get_ans(one,bit-1);
return min(get_ans(zero,bit-1),get_ans(one,bit-1))+(1<<bit);
}
int main()
{
int n;
cin >>n;
int x;
for(int i=0;i<n;i++) cin >>x,a.push_back(x);
cout <<get_ans(a,30)<<endl;
}
