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hdu 5014 Number Sequence（西安网络赛1008）

2024-07-22 19:02:31   230人阅读

Number Sequence

                                                                 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                              Total Submission(s): 633    Accepted Submission(s): 300
                                                                                                                                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
 
从大往小的找，知道发现n+j的值是二进制全为1的数，则会发现这一段（n-1）+（j+1)的值都为二进制全为1的数，这一段就处理好了，接着再往前处理。

代码：
//390ms
#include <iostream>
#include <cstdio>
#include <cstring>
const int maxn=100000+1000;
using namespace std;
int hash[20];
int a[maxn];
int b[maxn];
int main()
{
    int n;
    hash[0]=0;
    hash[1]=1;
    for(int i=1;i<19;i++)
    {
    hash[i+1]=(hash[i]<<1)+1;
    }
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++)
        scanf("%d",&a[i]);
        int temp;
        int sign=0;
        int m=n;
        while(m>=0)
        {
            if(m==0)//特判0
            {
                b[0]=0;
                break;
            }
           for(int i=0;i<=19;i++)//找到第一个各位全为1的且比m大
          {
            if(m<=hash[i])
            {
                temp=hash[i];
                break;
            }
          }
          for(int i=m-1;i>=0;i--)
          {
             if(m+i==temp)//找到左边对称的点
             {
                 sign=i;
                 break;
             }
          }
          for(int i=sign;i<=m;i++)
          {
              b[i]=temp-i;
          }
          m=sign-1;
        }
        long long ans=0;
        for(int i=0;i<=n;i++)
        ans=ans+(i^b[i]);
        printf("%I64d\n",ans);
        for(int i=0;i<n;i++)
        printf("%d ",b[a[i]]);
        printf("%d\n",b[a[n]]);
    }
    return 0;
}



hdu 5014 Number Sequence（西安网络赛1008）


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