HDU5014:Number Sequence

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

尽量错开对应位置的0,1即可，就是说将对应位的二进制按位取反了，当然不能用~符号
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
using namespace std;
#define ll __int64
#define up(i,x,y) for(i=x;i<=y;i++)
#define up2(i,x,y) for(i=x;y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define down2(i,x,y) for(i=x;y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define s(a) scanf("%d",&a)
#define s64(a) scanf("%I64d",&a)
#define w(a) while(a)

int n,a[100005],i,hash[100005];

int main()
{
    w(~s(n))
    {
        up(i,0,n)
        s(a[i]);
        printf("%I64d\n",(ll)n*n+n);
        mem(hash,-1);
        down(i,n,0)
        {
            if(hash[i]>=0) continue;
            int num=0,s=1,t=i;
            w(t)
            {
                int tem=((t&1)^1);
                num=num+s*tem;
                s*=2;
                t=t/2;
            }
            hash[num]=i;
            hash[i]=num;
        }
        printf("%d",hash[a[0]]);
        up(i,1,n)
        printf(" %d",hash[a[i]]);
        printf("\n");
    }

    return 0;
}