Number Sequence HDU

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a i ∈ 0,n0,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 = (a 0 � b 0) + (a 1 � b 1) +・・・+ (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 ≤ 10 5), The second line contains a 0,a 1,a 2,…,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 b 0,b 1,b 2,…,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

#include<map>#include<set>#include<queue>#include<stack>#include<vector>#include<math.h>#include<cstdio>#include<sstream>#include<numeric>//STL数值算法头文件#include<stdlib.h>#include <ctype.h>#include<string.h>#include<iostream>#include<algorithm>#include<functional>//模板类头文件using namespace std;typedef long long ll;const int maxn=110000;const int INF=0x3f3f3f3f;ll a[maxn];ll d[maxn];int main(){    ll n;    while(~scanf("%I64d",&n))    {        for(ll i=0; i<=n; i++)            scanf("%I64d",&a[i]);        memset(d,-1,sizeof(d));        ll ans=0;        for(ll i=n; i>=0; i--)        {            ll t=0;            if(d[i]==-1)            {                for(ll j=0;; j++)                {                    if(!(i&(1<<j)))  t+=(1<<j);                    if(t>=i)                    {                        t-=(1<<j);                        break;                    }                }                ans+=(i^t)*2;                d[i]=t;                d[t]=i;            }        }        printf("%I64d\n",ans);        for(ll i=0; i<=n; i++)            printf(i==n?"%I64d\n":"%I64d ",d[a[i]]);    }    return 0;}