HDU 5014 Number Sequence（异或 进制问题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5014

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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

思路：

尽可能找2^x-1，智商真是捉急啊！

最终可以全部异或为11111……（二进制位全是1）

所以最终的异或的和就是n*（n+1）；

代码如下：

#include <cstdio>#include <cstring>#define MAXN 100017typedef __int64 LL;int a[MAXN], vis[MAXN];int main(){    LL n;    while(~scanf("%I64d",&n))    {        memset(vis,-1,sizeof(vis));        for(int i = 0; i <= n; i++)        {            scanf("%d",&a[i]);        }        int m = 1;        while(m < n)//2^x-1        {            m = m*2+1;        }        for(int i = n; i >= 0; i--)        {            if(i <= m/2)                m/=2;            if(vis[i] == -1)            {                vis[i] = i^m;                vis[i^m] = i;            }        }        LL ans = n*(n+1);        printf("%I64d\n",ans);        for(int i = 0; i < n; i++)        {            printf("%d ",vis[a[i]]);        }        printf("%d\n",vis[a[n]]);    }    return 0;}

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

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online