2014西安网络赛1008||hdu5014 二进制

http://acm.hdu.edu.cn/showproblem.php?pid=5014

Problem Description

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a _i ∈ [0,n] 
● a _i ≠ a _j( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a ₀ ⊕ b ₀) + (a ₁ ⊕ b ₁) +···+ (a _n ⊕ b _n)

(sequence B should also satisfy the rules described above)

Now give you a number n and the sequence a. You should calculate the maximum integrating degree t and print the sequence b.

 

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line contains an integer n(1 ≤ n ≤ 10 ⁵), The second line contains a ₀,a ₁,a ₂,...,a _n.

 

Output

For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b ₀,b ₁,b ₂,...,b _n. There is exactly one space between b _i and b _i+1 (0 ≤ i ≤ n - 1). Don’t ouput any spaces after b _n.

 

Sample Input

   
   
   
   

    
    
    
    4
2 0 1 4 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    20
1 0 2 3 4
   
   
   
   

 

解题思路：我们知道一个数有n个二进制位，我们就把它异或成（1<< n）-1,就最大了，因此我们给每一个数从大到小用（1<< n）-1做异或就能得到与之对应的数，注意一定不能从小到大做。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

typedef __int64 LL;
const int N=100055;

int a[N],b[N],vis[N];
int n;

int get(int x)
{
    int num=0;
    while(x)
    {
        x>>=1;
        num++;
    }
    return num;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++)
            scanf("%d",&a[i]);
        memset(vis,-1,sizeof(vis));
        for(int i=n;i>=0;i--)
        {
            if(vis[i]!=-1)
                continue;
            int x=get(i);
            int tmp=((1<<x)-1)^i;
            b[i]=tmp;
            b[tmp]=i;
            //printf("%d\n",b[i]);
            vis[tmp]=1;
            vis[i]=1;
        }
        LL sum=0;
        for(int i=0;i<=n;i++)
            sum+=(LL)(i^b[i]);
        printf("%I64d\n",sum);
        for(int i=0;i<=n;i++)
            printf(i!=n?"%d ":"%d\n",b[a[i]]);
    }
    return 0;
}