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

     
     
     
     

      
      
      
       4
2 0 1 4 3 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
       20
1 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;
}