hdu5014——Number Sequence

Number Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 413    Accepted Submission(s): 199
Special Judge

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 = (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
   
   
   
   

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

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本题是特判题 ，所以只要找到任意的解就行了

0 1 2 3 4 5 ......n

依次用当前最大值去异或其他值，从那个位置开始从大到小填充，直到全部填完

另外要输出的最大值是有规律的，一个公差递增的数列

#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

int a[100010];
int b[100010];

int main()
{
	int n;
	while(~scanf("%d", &n))
	{
		for(int i = 0; i <= n; i++)
		{
			scanf("%d", &a[i]);
			b[i] = 0;
		}	
		int num = n;
		while( num >= 0 )
		{
			int pos = 0;
			int maxs = 0;
			int temp = -1;
			for(int i = 0; i <= num; i++)
			{
				temp = num ^ i;
				if(temp > maxs)
				{
					maxs = temp;
					pos = i;
				}
			}
			int idx = pos;
			while(b[idx] == 0 && idx <= n)
			{
				b[idx] = num;
				idx++;
				num--;
			}
		}
		printf("%I64d\n", (__int64)n * n + n);
		printf("%d", b[a[0]]);
		for(int i = 1; i <= n; i++)
			printf(" %d", b[a[i]]);
		printf("\n");
	}
	return 0;
}