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Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 150    Accepted Submission(s): 73
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

42 0 1 4 3

 

Sample Output

201 0 2 3 4

找二进制互补的数进行匹配

0      1      10      11      100      101      110      111      1000      1001      1010      1011      1100      1101      1110      1111      10000

0      1       2        3       4          5           6         7          8            9           10          11           12         13          14          15          16

可以发现，15和16互补，（14,1）（13,2）...也就是他们互补后为1111，即15，所以只要这样进行，就不会有浪费的1

#include<stdio.h>#include<algorithm>#include<string.h>#define N 100010using namespace std;typedef long long LL;LL i,j,k,m,n,x,y,z,sum;LL a[N],b[N];int main(){    while(scanf("%I64d",&n)!=EOF)    {        for(i=0;i<=n;i++)scanf("%I64d",&b[i]);        sum=0;        memset(a,0,sizeof(a));m=n;        while(n>0)        {            k=0;            for (i=20; i>=0; i--)                if ((n&(1<<i))!=0)                {                    k=1<<i;                    break;                }            sum+=(n-k+1)*(k*2-1)*2;            for (i=n;i+n>=(k*2-1);i--)                a[i]=k*2-1-i;            n-=(n-k+1)*2;        }        printf("%I64d\n",sum);        for (i=0;i<m;i++)printf("%I64d ",a[b[i]]);        printf("%I64d\n",a[b[m]]);    }    return 0;}