HDU - 5014 Number Sequence（贪心+位运算）

Number Sequence

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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

 42 0 1 4 3 

 

Sample Output

 201 0 2 3 4 

 

题目大意：
有一个a集合满足
● a _i ∈ [0,n] 
● a _i ≠ a _j( i ≠ j ) 

同样的b也满足该规则，
现在问你是否存在一个序列b，使得t最大

t = (a ₀ � b ₀) + (a ₁ � b ₁) +・・・+ (a _n � b _n)

n(1 ≤ n ≤ 10 ⁵)

解析：贪心题，但是比赛时候我用的是回溯求解该题，一直超时。后来想想我太笨了如果用回溯的话复杂度为O(10^10)，太大了。

其实弄清楚了这题的思路敲起来还是蛮简单的，每一个数的最大的异或是和他互补的数。如10101^01010最大，如果全部都取最大，那么最后的t肯定最大。

那么一个数的互补怎么求呢？ 就是这个数 异或上 与该数二进制位数相同，且都为1的数

如

       1010

xor  1111

       0101

弄清楚这个就好办了。求出所有数的互补，把它存入一个数组中，最后按照输入时的顺序输出该数组，就好了。

注意：Hdu有个比较坑的地方，就是不能开成long long 所以应该把t 声明为__int64才能过，坑爹啊。

#include<cstdio>#include<cstring>#include<cmath>using namespace std;typedef __int64 LL;const int MAX = 100005;int a[MAX],ans[MAX],vis[MAX];int main() {int n;while(scanf("%d",&n)!=EOF) {for(int i = 0;i <= n;++i)scanf("%d",&a[i]);memset(vis,0,sizeof(vis));memset(ans,0,sizeof(ans));for(int i = n;i > 0;--i) {if(vis[i])continue;int d = log2(i)+1;int temp = ((1 << d) - 1)  ^ i;ans[i] = temp;ans[temp] = i;vis[temp] = vis[i] = 1;}LL sum = 0;for(int i = 0;i <= n;++i)sum += (LL)(i ^ ans[i]);printf("%I64d\n",sum);for(int i = 0;i < n;++i)printf("%d ",ans[a[i]]);printf("%d\n",ans[a[n]]);}return 0;}