HDU 5014 Number Sequence)

Number Sequence Time Limit: 4000/2000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)
Total submission (s): 545 accepted submission (s): 258
Special Judge

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

● AI in [0, N]
● AI = AJ (I = J)

For sequence a and sequence B, the integrating Degree t is defined as follows ("writable" denotes exclusive or ):

T = (A0 rjb0) + (A1 rjb1) + · + (an 1_bn)

(Sequence B shoshould also satisfy the rules described abve)

Now give you a number N and the sequence A. You shoshould calculate the maximum integrating Degree t and print the sequence B.
 
Inputthere 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,...,.
 
Outputfor 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

42 0 1 4 3

 
Sample output

201 0 2 3 4

 
Source2014 ACM/ICPC Asia Regional Xi 'an online

Question and question:

The result is N * (n + 1). You can see the first five. The rest of the output goes down from the very beginning, and 2 ^ (LEN)-1-I can be output. The Edge tag is calculated by the side (Len is the number of digits in the binary representation of the current number ).

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int len(int n){    int ans=0;    while(n)    {        n>>=1;        ans++;    }    return ans;}int s[100010];int t[100010];int main(){    int n,le,v;    __int64 d;    while(scanf("%d",&n)!=EOF)    {        memset(s,-1,sizeof(s[0])*(n+5));        for(int i=0;i<=n;i++)        {            scanf("%d",&t[i]);        }        for(int i=n;i>=0;i--)        if(s[i]<0)        {            le=len(i);            v=1<<le;            v--;            s[i]=v-i;            s[v-i]=i;        }        __int64 m=(__int64)n;        d=m*(m+1);        printf("%I64d\n",d);        for(int i=0;i<n;i++)        {            printf("%d ",s[t[i]]);        }        printf("%d\n",s[t[n]]);    }    return 0;}

HDU 5014 Number Sequence)