HDU 5014解题报告

Number Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536
K (Java/Others)

Total Submission(s): 1279 Accepted Submission(s): 548

Special Judge

Problem Description

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● ai ∈ [0,n]

● ai ≠ aj( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)

(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,...,an.



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,...,bn. There is exactly one space between bi and bi+1(0
≤ i ≤ n - 1). Don’t ouput any spaces after bn.



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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解题思路

这道题关键是在a[i]的范围是在0~n之间，而要保证a[0]^b[0]+a[1]^b[1]+……a
^b
的值最大。也就是要尽量将a[i]在二进制下的每一位取反。但是这样容易产生重复的问题，不知道如何去取舍。比如样例中的a[i]为2时，b[i]可以取1，a[i]为0时，b[i]可以取1。这样便产生了问题。。所以为了保证数值的最大，2^1=3，而0^1=1,所以应该优先满足较大的数。又因为a[i]和b[i]的范围都是0~n。所以可以从n到0做一个循环，从大到小逐个数进行考虑。将这个i的每一位取反，得到一个新的数。然后就得到一对数，保证他们局部的异或值最大。然后将这些异或的值加起来即可。

总的来说，这道题用到了贪心的思想，而且用到了哈希的思想，建立一一映射。还是比较巧妙的。

参考代码

#include<cstdio>
#include<iostream>
#define __CLR(x) memset(x,-1,sizeof(x))
#define ll long long
using namespace std;
const int maxn=100010;

int a[maxn],b[maxn];
int main()
{
    //ios_base::sync_with_stdio(0);
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<=n; i++)
            scanf("%d",&a[i]);
        ll ans=0;
        __CLR(b);
        for(int i=n; i>=0; i--)
        {
            if(b[i]>=0) continue;
            int num=0,k=i,s=1;
            while(k>0)
            {
                int t=(k%2)?0:1;
                num=num+s*t;
                s*=2;
                k/=2;
            }
            b[num]=i;
            b[i]=num;
            ans+=2*(b[i]^b[num]);
        }
        printf("%I64d\n",ans);
        for(int i=0;i<=n;i++)
        {
            if(i>0) printf(" ");
            printf("%d",b[a[i]]);
        }
        printf("\n");
    }
}