Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 697 Accepted Submission(s): 332
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
随意写一个二进制数x=100100,总能找到唯一的一个数与它异或得到111111；对于每个N位二进制数，每次总使它与一个数异或得到11...11（N个1）。即为最优解，且唯一。
#include"stdio.h"
#include"math.h"
#include"string.h"
#define LL __int64
#define N 100005
int a[N],p[N],b[N];
int fun(int x)
{
int t=1;
while(t<=x)
t*=2;
return x^(t-1);
}
int main()
{
int i,n;
while(scanf("%d",&n)!=-1)
{
for(i=0;i<=n;i++)
scanf("%d",&a[i]);
memset(p,-1,sizeof(p));
for(i=n;i>=0;i--)
{
if(p[i]!=-1)
continue;
int t=fun(i);
p[t]=i;
p[i]=t;
}
LL s=0;
for(i=0;i<=n;i++)
{
b[i]=p[a[i]];
s+=a[i]^b[i];
}
printf("%I64d\n",s);
for(i=0;i
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