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 = (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
   
   
   
   

 

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]);
			else
				printf(" %d",data[tem]);
		}
		puts("");
	}
	return 0;
}