Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2075 Accepted Submission(s): 814
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
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=10e5+10;
const int inf=(1<<30);
int arr[maxn];
int res[maxn];
bool vis[maxn];
int main()
{
int n;
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
for(int i=0;i<=n;i++)
scanf("%d",&arr[i]);
__int64 sum=0;
for(int i=n;i>=0;i--)
{
int t,k=1;
if(vis[i]) continue;
for(int j=0;;j++)
{
t=(k<<j)-1;
if(i<=t) break;
}
res[i]=t-i;
res[t-i]=i;
vis[i]=vis[t-i]=1;
sum+=2*(i^(t-i));
}
printf("%I64d\n",sum);
for(int i=0;i<n;i++)
printf("%d ",res[arr[i]]);
printf("%d\n",res[arr[n]]);
}
return 0;
}