hdu 5014 Number Sequence 位运算+规律

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

 

题意：输入n，然后输入n+1个数，是【0，n】中不同的数。然后再找到一种排列，也是【0，n】中不同数组成的；

然后一一对应，异或后相加；输入结果 和新数列

思路：

从题目中的

2 0 1 4 3

1 0 2 3 4  看出每个数都可以找到完全匹配的，异或后不浪费任何一个1（也就是1所在位均和0匹配）。

再手算下

0 1 2 3 4 5 

5 2 1 4 3 0

0 1 2 3 4 5 6

0 6 5 4 3 2 1

匹配后都能符合这个规律。没有1浪费掉。

所以sum=(0+n)*(n+1)/2*2=n*(n+1)

然后找和 i 匹配的；

如 i=5 = 101(B)  用111(B)和他^异或操作，就能得到相反数10(B)=2；

如 i=7 = 111(B) 也用111(B)，^后，得到相反数0(B)=0；

要保证111。。。 的位数不能比i 的第一个1所在的位数大；

#include<stdio.h>#include<string.h>int data[200000];int main(){int n,i,tem,yi;while(scanf("%d",&n)!=EOF){ memset(data,-1,sizeof(data));yi=1;while(yi<=n){yi=yi*2+1;}for(i=n;i>=0;i--){if(yi/2>=i){yi>>=1;}if(data[i]==-1){data[i]=yi^i;data[yi^i]=i;}} printf("%I64d\n",(__int64)(0+n)*(n+1));for(i=0;i<n+1;i++){scanf("%d",&tem);if(i==0)printf("%d",data[tem]);elseprintf(" %d",data[tem]);}puts("");}return 0;}