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

                                                                   

题意：

给出一列数只包括0~n （1<=n<=10^5）求出一个排列使得一一对应的异或值总和最大~

思路：

可以找到其规律，先写出0~10 的二进制数

0  0000 ，1  0001，2  0010,3  0011,4  0100， 5  0101，6 0110，7  0111，8  1000,9  1001， 10  1010

可以看出倘若每一个数所占的位数都为1 的话，那么得到的异或值就是最大的，而刚好得到的最大异或值减去原来的数就是得到的配对的数。

code:

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;int a[100005], ans[100005],n;int cal(int n){    int l = 0;    while(n){        n >>= 1;        l ++;    }    return l;}int main(){    //freopen("in", "r", stdin);    while(~scanf("%d", &n)){        memset(ans, -1, sizeof(ans));        for(int i = 0; i <= n; i ++){            scanf("%d", &a[i]);        }        ll num,t,sum = 0;        for(int i=n; i>=0; i--){            if(ans[i] == -1){                int len=cal(i);                num = (1<<len) - 1;                t = num^i;                if(ans[t] == -1){                    ans[i] = t;                    ans[t] = i;                    sum += (num*2);                }            }        }        printf("%d\n", sum);        printf("%d", ans[a[0]]);        for(int i=1;i<=n;i++){            printf(" %d", ans[a[i]]);        }        printf("\n");    }    return 0;}