The HDU-5014-Number sequence.

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

● AI in [0, N]
● AI = AJ (I = J)

For sequence a and sequence B, the integrating Degree t is defined as follows ("writable" denotes exclusive or ):

T = (A0 rjb0) + (A1 rjb1) + · + (an 1_bn)

(Sequence B shoshould also satisfy the rules described abve)

Now give you a number N and the sequence A. You shoshould calculate the maximum integrating Degree t and print the sequence B.
 
Inputthere 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,...,.
 
Outputfor 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

42 0 1 4 3

 
Sample output

201 0 2 3 4

 
Source2014 ACM/ICPC Asia Regional Xi 'an online
Idea: Start from the largest number to find the number that can be paired to make them unique or the maximum value. Fortunately, I didn't hit the game because Longlong wa had several rounds.

#include <stdio.h>int num[100005],d[100005];int main(){    int n,i,j,t;    long long ans;    while(~scanf("%d",&n))    {        for(i=0;i<=n;i++) d[i]=-1;        for(i=0;i<=n;i++) scanf("%d",&num[i]);        ans=0;        for(i=n;i>=0;i--)        {            if(d[i]==-1)            {                t=0;                for(j=0;;j++)                {                    if(!(i&(1<<j))) t+=1<<j;                    if(t>=i) break;                }                t-=(1<<j);                ans+=(i^t)+(i^t);                d[i]=t;                d[t]=i;            }        }        printf("%I64d\n",ans);        for(i=0;i<n;i++) printf("%d ",d[num[i]]);        printf("%d\n",d[num[n]]);    }}

The HDU-5014-Number sequence.