HDU-5014-Number Sequence

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

 

思路：从最大的一个数开始找能配对使他们的异或值最大的一个数即可。幸亏比赛的时候不是我敲，因为longlong WA 了好几发。

#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]]);    }}