HDU-5014-Number Sequence - c++编程基础

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
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 
  
   

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<
   
    =i) break; } t-=(1<