首页 > 代码库 > HDU-5014-Number Sequence

HDU-5014-Number Sequence

2024-07-22 14:19:26   226人阅读

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
 

思路：从最大的一个数开始找能配对使他们的异或值最大的一个数即可。幸亏比赛的时候不是我敲，因为longlong WA 了好几发。

#include <stdio.h>

int num[100005],d[100005];

int main()
{
    int n,i,j,t;
    long long ans;

    while(~scanf("%d",&n))
    {
        for(i=0;i<=n;i++) d[i]=-1;

        for(i=0;i<=n;i++) scanf("%d",&num[i]);

        ans=0;

        for(i=n;i>=0;i--)
        {
            if(d[i]==-1)
            {
                t=0;

                for(j=0;;j++)
                {
                    if(!(i&(1<<j))) t+=1<<j;

                    if(t>=i) break;
                }

                t-=(1<<j);

                ans+=(i^t)+(i^t);

                d[i]=t;
                d[t]=i;
            }
        }

        printf("%I64d\n",ans);

        for(i=0;i<n;i++) printf("%d ",d[num[i]]);

        printf("%d\n",d[num[n]]);
    }
}


HDU-5014-Number Sequence


联系
我们