动态规划——Compatible Numbers

Two integers x and y are compatible, if the result of their bitwise "AND" equals zero, that is, a & b = 0. For example, numbers 90 (1011010₂) and 36 (100100₂) are compatible, as 1011010₂ & 100100₂ = 0₂, and numbers 3 (11₂) and 6 (110₂) are not compatible, as 11₂ & 110₂ = 10₂.

You are given an array of integers a₁, a₂, ..., a_n. Your task is to find the following for each array element: is this element compatible with some other element from the given array? If the answer to this question is positive, then you also should find any suitable element.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁶) — the number of elements in the given array. The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 4·10⁶) — the elements of the given array. The numbers in the array can coincide.

Output

Print n integers ans_i. If a_i isn't compatible with any other element of the given array a₁, a₂, ..., a_n, then ans_i should be equal to -1. Otherwise ans_i is any such number, that a_i & ans_i = 0, and also ans_i occurs in the array a₁, a₂, ..., a_n.

Example

Input

290 36

Output

36 90

Input

43 6 3 6

Output

-1 -1 -1 -1

Input

510 6 9 8 2

Output

-1 8 2 2 8

题意：给n个数，对于某个数a[i]如果有a[j]使a[i]&a[j]==0，则对于a[i]对应a[j]，否则对应-1。要求输出每个数所对应的数。如果有多种情况输出其中一种即可。

思路：最容易想到的是对每个数在数组中遍历寻找，但是无疑会TLE。。。

于是看数据知道a[i]最大为4·10⁶

^{那么可以设置一个常数k=(1<<22)-1；然后用dp[i^a[i]]=a[i]保存a[i]和与k的异或的值的对应关系。}

之后寻找时对每个位枚举即可。

详见代码。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int k=(1<<22)-1;int n;int dp[k+1];int a[4000005];int main(){    while(~scanf("%d", &n))    {        memset(a, 0, sizeof(a));        memset(dp, 0, sizeof(dp));        for(int i=1; i<=n; i++)        {            scanf("%d", &a[i]);            dp[k^a[i]]=a[i];        }        for(int i=k; i>=0; i--)        {            if(!dp[i])             //如果没有这个状态就不用判断                for(int j=0; j<22; j++)                {                    if(dp[i|(1<<j)])             //枚举每一位是否出现过这种状态                        dp[i]=dp[i|(1<<j)];                }        }        for(int i=1; i<=n; i++)        {            if(dp[a[i]])                printf("%d", dp[a[i]]);            else                printf("-1");            if(i==n)                printf("\n");            else                printf(" ");        }    }    return 0;}