找规律

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int n,a[100100],sig[100100];
long long int ans;

int wei(int x)
{
    if(x==0) return 0;
    return log(x*1.)/log(2.0);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        memset(sig,-1,sizeof(sig));
        for(int i=n;i>=0;i--)
        {
            if(n%2==0&&i==0)
            {
                sig[0]=0;
                continue;
            }
            if(sig[i]!=-1)
            {
                ans+=i^sig[i];
                continue;
            }
            int w=wei(i);
            w++;
            int fan=((1<<w)-1)^i;
            sig[i]=fan;
            sig[fan]=i;
            ans+=i^sig[i];
        }
        printf("%I64d\n",ans);
        for(int i=0;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            if(i) putchar(32);
            printf("%d",sig[x]);
        }
        putchar(10);
    }
    return 0;
}