hdu 5014 亦或的性质

根据亦或的性质，要获得最大的和，那么两两配对的数的二进制数的每一位都不相同，所以会形成一种对称的局势．

比如１１１１到００００之间就是两两对称

1010和０１０１之间也是两两对称，

根据这种性质，那么只要利用hash提前预处理出配对的数，然后求解就好了

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#define MAX 300007

using namespace std;

int n;
int a[MAX],b[MAX];
int hash[MAX];

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        for ( int i = 0 ; i <= n ; i++ ) scanf ( "%d" , &a[i] );
        memset ( hash , 0 , sizeof ( hash ) );
        int temp = n;
        int total = 1;
        hash[0] = 0;
        while ( temp > 0 )
        {
            total = 1;
            while ( total <= temp ) total<<=1;
            total--;
            int next = (total^temp)-1;
            while ( 1 )
            {
                hash[temp^total] = temp;
                hash[temp] = temp^total;
                if ( (temp^total)+1 == temp ) break;
                temp--;
            }
            temp = next;
        }
        long long ans = 0;
        for ( int i = 0 ; i <= n ; i++ ) 
        {
            b[i] = hash[a[i]];
            ans += b[i]^a[i];
        }
        printf ( "%I64d\n" , ans );
        for ( int i = 0 ; i < n ; i++ ) printf ( "%d " , b[i] );
        printf ( "%d\n" , b[n] );
    }
}