AcWing 143. 最大异或对
原题链接
简单
作者:
纪年
,
2021-05-11 11:28:47
,
所有人可见
,
阅读 341
#include <iostream>
using namespace std;
const int N = 100010;
int n;
int son[31 * N][2], idx;
void insert(int x)
{
int p = 0;
for (int i = 30; i >= 0; i -- )
{
int u = x >> i & 1;
if (!son[p][u]) son[p][u] = ++ idx;
p = son[p][u];
}
}
int query(int x)
{
int p = 0, res = 0;
for (int i = 30; i >= 0; i -- )
{
int u = x >> i & 1;
if (son[p][!u])
{
res = res * 2 + 1;
p = son[p][!u];
} else{
res = res * 2;
p = son[p][u];
}
}
return res;
}
int main()
{
int ans = 0;
scanf("%d", &n);
while (n -- )
{
int x;
scanf("%d", &x);
insert(x);
ans = max(ans, query(x));
}
printf("%d\n", ans);
return 0;
}