题干：

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 bn. 

Sample Input

4
2 0 1 4 3

Sample Output

20
1 0 2 3 4

题目大意：

● a i ∈ [0,n] 
● a i ≠ a j( i ≠ j ) 

同样的b也满足该规则，
现在问你是否存在一个序列b，使得t最大

t = (a 0 异或 b 0) + (a 1 异或 b 1) +・・・+ (a n 异或 b n)

n(1 ≤ n ≤ 10 5)

解题报告：

   直接从大到小贪心，让每一个数和 他按位取反的那个数匹配。这样虽然可能会剩下一个0，但是易知这样一个1都没有浪费，所以这样得到的结果一定是最优的。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,a[MAX];
int b[MAX];
ll ans;
int main()
{
  while(cin>>n) {
    ans = 0;
    memset(b,-1,sizeof b);b[0]=0;
    for(int i = 0; i<=n; i++) cin>>a[i];
    for(int sta = n; sta>0; sta--) {
      if(b[sta] != -1) continue;
      int bit;
      for(bit = 22; bit>=0; bit--) {
        if(sta&(1<<bit)) break;
      }
      if(bit == -1)continue;
      int oth = 0;  
      for(int i = bit; i>=0; i--) {
        if(sta & (1<<i)) continue;
        else oth |= (1<<i);
      }
      b[oth] = sta; b[sta] = oth;
    }
    for(int i = 0; i<=n; i++) ans += i^b[i];
    printf("%lld\n",ans);
    for(int i = 0; i<=n; i++) printf("%d%c",b[a[i]],i == n ? '\n' : ' ');   
  }
  
  return 0 ;
}