2014西安赛区网络赛 5014 Number Sequence

Number Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 214 Accepted Submission(s): 110

Special Judge

Problem Description

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

● ai ∈ [0,n]

● ai ≠ aj( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)

(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,...,an.



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,...,bn. There is exactly one space between bi and bi+1(0
≤ i ≤ n - 1). Don’t ouput any spaces after bn.



Sample Input

4
2 0 1 4 3

Sample Output

20
1 0 2 3 4

给定n 数组a中保存0--n n+1个数 每个数都不一样
数组b中同样存储这n+1个数 问 b数组中的数应该如何安排使得 ai^bi的和最大
暴力模拟 对每位取反 去异或 求最大值

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>

#define eps 1e-8
#define op operator
#define MOD  10009
#define MAXN  100010

#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define FOV(i,a,b)  for(int i=a;i>=b;i--)
#define REP(i,a,b)  for(int i=a;i<b;i++)
#define REV(i,a,b)  for(int i=a-1;i>=b;i--)
#define MEM(a,x)    memset(a,x,sizeof a)
#define ll __int64
using namespace std;

ll n;
ll a[MAXN],b[MAXN];

ll dfs(ll x)
{
    if(x<=0)  return 0;
    ll y=x;
    ll m=1;
    while(y>1)
    {
        y/=2;
        m*=2;
    }
//    cout<<m<<endl;
    ll res=0;
    ll q=m-1;
    for(ll i=m;i<=x;i++)
    {
        res+=(i^q)*2;
        b[i]=q; b[q]=i;
        q--;
    }
    res+=dfs(q);
    return res;
}

int main()
{
//freopen("ceshi.txt","r",stdin);

    while(scanf("%I64d",&n)!=EOF)
    {
        for(ll i=0;i<=n;i++)
            scanf("%I64d",&a[i]);
        MEM(b,0);
//        int res;
        printf("%I64d\n",dfs(n));
        printf("%I64d",b[a[0]]);
        for(ll i=1;i<=n;i++)
            printf(" %I64d",b[a[i]]);
        puts("");
    }
    return 0;
}