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

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：

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 100017
typedef __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

4
2 0 1 4 3

Sample Output

20
1 0 2 3 4

Source

2014 ACM/ICPC Asia Regional Xi‘an Online