CodeForces 165E 【状压DP】

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 原题链接

 

E. 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, numbers90(1011010₂) and36(100100₂) are compatible, as1011010₂&100100₂ = 0₂, and numbers3(11₂) and6(110₂) are not compatible, as11₂&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 containsn 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 arraya₁, a₂, ..., a_n, thenans_i should be equal to -1. Otherwiseans_i is any such number, thata_i&ans_i = 0, and alsoans_i occurs in the arraya₁, a₂, ..., a_n.

Examples

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：n个正整数，对于每一个ai，是否存在aj&ai=0，若存在则输出aj否则输出-1；
            脑洞题，dp[N^a[i]]=a[i];（N=1<<22-1） 往下顺推，具体看代码。

#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;int dp[1<<22],a[1001000],c[25],n;int main(){    c[0]=1;    for(int i=1;i<=23;i++)        c[i]=c[i-1]<<1;    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            dp[a[i]]=-1;        }        for(int i=1;i<=n;i++)        dp[(c[22]-1)^a[i]]=a[i];        for(int i=c[22]-1;i>0;i--)        if(dp[i]>0)        {            for(int j=0;j<22;j++)                if(i&c[j])            {                dp[i-c[j]]=dp[i];            }        }        for(int i=1;i<n;i++)            printf("%d ",dp[a[i]]);        printf("%d\n",dp[a[n]]);    }    return 0;}