Codeforces Round #613 (Div. 2) 异或最大值最小(分治)

Dr. Evil Underscores

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).

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

题意:寻找x与数组a异或,求异或后数组内最大值,使得最大值最小。

题解:分治 字典树 dp(dp待补)

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
using ll = long long;
ll solve(vector<ll> a,int b)
{
    if(b==-1)return 0;
    ll temp = 1ll<<b;
    vector<ll> l, r;
    for(auto i:a)
	{
        if(i<temp)l.push_back(i);
        else r.push_back(i-temp);
    }
    if(l.empty())return solve(r,b-1);
    if(r.empty())return solve(l,b-1);
    return min(solve(l,b-1),solve(r,b-1))+temp;
}
int main(){
    int n;
    cin>>n;
    vector<ll> a(n);
    for(int i=0;i<n;i++)
	cin>>a[i];
    sort(a.begin(),a.end());
    cout<<solve(a,32)<<endl;
    return 0;
}
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转载自blog.csdn.net/qq_43381887/article/details/104074071