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

题解及题解：

瞎搞就搞出来了，最后的结果是n*(n+1)，手算一下前5个就能看出来。剩下的输出从大开始向下遍历，输出2^(len)-1-i就可以了，边计算边标记（len为当前数的2进制表示形式的位数）。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int len(int n)
{
    int ans=0;
    while(n)
    {
        n>>=1;
        ans++;
    }
    return ans;
}

int s[100010];
int t[100010];
int main()
{
    int n,le,v;
    __int64 d;
    while(scanf("%d",&n)!=EOF)
    {
        memset(s,-1,sizeof(s[0])*(n+5));
        for(int i=0;i<=n;i++)
        {
            scanf("%d",&t[i]);
        }

        for(int i=n;i>=0;i--)
        if(s[i]<0)
        {
            le=len(i);
            v=1<<le;
            v--;
            s[i]=v-i;
            s[v-i]=i;
        }
        __int64 m=(__int64)n;
        d=m*(m+1);
        printf("%I64d\n",d);
        for(int i=0;i<n;i++)
        {
            printf("%d ",s[t[i]]);
        }
        printf("%d\n",s[t[n]]);
    }
    return 0;
}