Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 561 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 4 2 0 1 4 3
Sample Output 20 1 0 2 3 4

這題在之前做時候完全想不到思路，直到比賽2小時腦子裡突然閃現出一個YY思路，那就是如果對於一個0-n的序列，能構成的最大異或序列，需要這樣構想，對於n，能構成的最大數字需要與n取反配對（n與n取反異或運算後所有二進位制位置均為1，即最大），我們設n取反為m，那麼n-1應當與m+1配對，依次類推。這樣，我們得到了m-n所有的配對方式。然後，不斷迴圈配對m-1與m-1取反區間內的數字，直到配對至0則完成全部配對，此時即為異或配對的最大值。。。輸出即為答案。。。還是挺簡單的嘛。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
using namespace std;
__int64 a[100005],pa[100005];
__int64 cal(__int64 pa)
{
    __int64 ans=0,n=0;
    while(pa>0)
    {
        a[n]=pa%2;
        n++;
        pa=pa/2;
    }
    __int64 tap=1;
    for(int j=0;j<n;j++)
    {
        if(a[j]==0)
            ans+=tap;
        tap=tap*2;
    }
    return ans;
}
int main()
{
 //   freopen("in.txt","r",stdin);
    int n;
    while(cin>>n)
    {
        for(int i=0;i<=n;i++)
            scanf("%d",&pa[i]);
        __int64 pat=cal(n),sum=0,psum=0;
        sum+=(n-pat)*(pat+n+1);
        for(int i=pat,j=n;i<=n;i++)
        {
            a[i]=j;
            psum+=(i^j);
            j--;
        }
        pat--;
        while(pat>0)
        {
            __int64 item=cal(pat);
            for(int i=item,j=pat;i<=pat;i++)
            {
                a[i]=j;
                psum+=(i^j);
                j--;
            }
            sum+=(pat-item)*(pat+item+1);
            pat=item;
            pat--;
        }
        cout<<psum<<endl;
        for(int i=0;i<n;i++)
            printf("%d ",a[pa[i]]);
        printf("%d\n",a[pa[n]]);
    }
    return 0;
}