HDU -- 5014 Number Sequence

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
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题意：
给出的 0 - n 之间的数字，求出它们异或的和的最大；
任意的数字 ，例如 0 1 0 1，以其异或的值最大是 1 0 1 0；
因此 我们就 可以通过 0 1 0 1 异或 1 1 1 1 得到 1 0 1 0；
那么就可以求出各自对应的值；

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#define ll __int64#define N 100005using namespace std;int a[N];int vis[N];int b[N];int main() {    int n;    while (scanf("%d",&n) != EOF) {        memset(a,0,sizeof(a));        memset(vis,0,sizeof(vis));        memset(b,0,sizeof(b));        for (int i = 0; i <= n; i++)            scanf("%d",&a[i]);        for (int i = n; i > 0; i--) {  // 逆序            if (vis[i] == 0 ) {                int d = log2(i) + 1 ; // 为了求出求位数，例如 i = 8 ；d = 4

                int t1 = ((1 << d) - 1);// 位数减去1，可以得到该位数的11……1； t1 = 15 ；（111）

                int t = t1 ^ i; //求出与他异或的值                b[i] = t;                b[t] = i;                vis[i] = 1;                vis[t] = 1;            }        }        ll sum = 0;        for (int i = 0; i <= n; i++)             sum += i ^ b[i];        printf("%I64d\n",sum);        for (int i = 0; i < n; i++)            printf("%d ",b[a[i]]);        printf("%d\n",b[a[n]]);    }}