题意：
这个题目的意思是给出0~n的排列，然后找出与这个序列的配对使(a0 ? b0) + (a1 ? b1) +?+ (an ? bn)最大。。
思路：
从大到小遍历每个数，然后找到与这个数二进制位数互补的数，那么他们的抑或值必定是pow（2，n）-1,，肯定是最大的。。。。

题目：

Number Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 713 Accepted Submission(s): 337
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 = (a ₀ ? b ₀) + (a ₁ ? b ₁) +???+ (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 ≤ 10 ⁵), The second line contains a ₀,a ₁,a ₂,...,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 b ₀,b ₁,b ₂,...,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

Source 2014 ACM/ICPC Asia Regional Xi'an Online 代码：

#include
  
   
#include
   
     #include
    
      #include
     
       #include
       #include
       
         #include
        
          #include
         
           #include
          
            #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f using namespace std; const int maxn=100000+10; int a[maxn],n,ans[maxn]; bool vis[maxn]; int gao(int k) { int cnt=0; while(k) { k>>=1; cnt++; } return cnt; } int main() { while(~scanf("%d",&n)) { memset(vis,false,sizeof(vis)); for(int i=0;i<=n;i++) scanf("%d",&a[i]); for(int i=n;i>=0;i--) { if(vis[i]) continue; int move=gao(i); int temp=(((1<

hdu5014 Number Sequence（异或运算）

Number Sequence