Codeforces Round #673 (Div. 2)[字典树]E.xor inverse
code:
#include<iostream>
using namespace std;
const int N = 3e5 + 5;
typedef long long ll;
int trie[N * 30][2];
int count = 0;
ll sz[N * 30]={0};
ll dp[35][2]={0};
void insert(int x){
int root = 0;
for(int i = 30;i >= 0;i--){
int j =(x>&g.
原创
2020-10-11 11:33:21 ·
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