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

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Special Judge

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 = (a₀ ⊕ b₀) + (a₁ ⊕ b₁) +···+ (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 ≤ 10⁵), The second line contains a₀,a₁,a₂,...,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 b₀,b₁,b₂,...,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

Source

2014 ACM/ICPC Asia Regional Xi‘an Online

思路： 网上已经有很多解题报告了，解法都差不多，给定一个数n，求这n+1个数与另外n+1个数相匹配求最大的异或值之和；

可以发现规律： 从n开始遍历到0，让每个值都找到另一个“互补”的值，比如 ：n=9,那么就有10个数，最大的是9，其二进制表示为1001，那么它就应该与0110匹配~

（一定得从最大值n开始遍历，才符合贪心的正确性）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cstdio>
#include <cmath>
const int N=1e5+100;
using namespace std;
int hash[N];
int a[N];

int main()
{
    int n;
    while(cin>>n)
    {
      for(int i=1;i<=n+1;i++)
      {
        scanf("%d",&a[i]);
      }
      memset(hash,-1,sizeof(hash));
      for(int i=n;i>=0;i--) //将n个数从大到小排列进行位处理，使得“互补”，这样异或值才是最大的;
      {
       if(hash[i]>-1)continue;
       int sum=0,cnt=1,s=i;
       while(s)
       {
         int t=(s&1)^1;
         sum+=t*cnt;
         cnt*=2;
         s/=2;
       }

       hash[i]=sum;
       hash[sum]=i;
      }
      printf("%I64d\n",(long long )n*n+n);
      for(int i=1;i<=n+1;i++)
      {
        if(i==1)printf("%d",hash[a[i]]);
        else printf(" %d",hash[a[i]]);
      }
      printf("\n");
    }
    return 0;
}