HDOJ 5014 Number Sequence

找规律

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 303    Accepted Submission(s): 149
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

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int n,a[100100],sig[100100];long long int ans;int wei(int x){    if(x==0) return 0;    return log(x*1.)/log(2.0);}int main(){    while(scanf("%d",&n)!=EOF)    {        ans=0;        memset(sig,-1,sizeof(sig));        for(int i=n;i>=0;i--)         {            if(n%2==0&&i==0)             {                sig[0]=0;                continue;            }            if(sig[i]!=-1)            {                ans+=i^sig[i];                continue;            }            int w=wei(i);                w++;            int fan=((1<<w)-1)^i;            sig[i]=fan;            sig[fan]=i;            ans+=i^sig[i];        }        printf("%I64d\n",ans);        for(int i=0;i<=n;i++)        {            int x;            scanf("%d",&x);            if(i) putchar(32);            printf("%d",sig[x]);        }        putchar(10);    }    return 0;}