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

随意写一个二进制数x=100100,总能找到唯一的一个数与它异或得到111111；对于每个N位二进制数，每次总使它与一个数异或得到11...11（N个1）。即为最优解，且唯一。

#include"stdio.h"
#include"math.h"
#include"string.h"
#define LL __int64
#define N 100005
int a[N],p[N],b[N];
int fun(int x)
{
    int t=1;
    while(t<=x)
       t*=2;
    return x^(t-1);
}
int main()
{
    int i,n;
    while(scanf("%d",&n)!=-1)
    {
        for(i=0;i<=n;i++)
            scanf("%d",&a[i]);
        memset(p,-1,sizeof(p));
        for(i=n;i>=0;i--)
        {
            if(p[i]!=-1)
                continue;
            int t=fun(i);
            p[t]=i;
            p[i]=t;
        }
        LL s=0;
        for(i=0;i<=n;i++)
        {
            b[i]=p[a[i]];
            s+=a[i]^b[i];
        }
        printf("%I64d\n",s);
        for(i=0;i<n;i++)
            printf("%d ",b[i]);
        printf("%d\n",b[n]);
    }
    return 0;
}