HDU5015~Number Sequence（位运算）

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2516    Accepted Submission(s): 1010
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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

Recommend

hujie

 

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题意：n+1个数的序列，每个数不重复，都在0~n范围内，求最大的t，公式题中已给出，就是保证每个a[i]^b[i]最大，a[i]^b[i]最大，也就是转换成二进制，每一位都不同，则对每一个a[i]都有一个唯一的b[i]使得a[i]^b[i]最大，暴力，每次把1往左移一位，移后判断一下&a[i]==0，如果=0，则加上，以此求出最大。

#include<stdio.h>#include<string.h>#include<math.h>typedef  long long ll;using namespace std;ll a[100005];ll b[100005];int main(){   ll n;   while(~scanf("%lld",&n))   {       memset(b,-1,sizeof(b));       for(int i=0;i<=n;i++)        scanf("%lld",&a[i]);        ll ans=0,p;       for(int i=n;i>=0;i--)       {           p=0;           if(b[i]==-1)           {               for(int j=0;(1<<j)<i;j++)                  if((i&(1<<j))==0)                    p+=(1<<j);               b[p]=i;               b[i]=p;               ans+=(i^p)*2;           }       }       printf("%lld\n",ans);       for(int i=0;i<=n;i++)       {           if(i==0)            printf("%lld",b[a[i]]);           else            printf(" %lld",b[a[i]]);       }       printf("\n");   }}