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 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

Source 2014 ACM/ICPC Asia Regional Xi'an Online

题解及题解：

瞎搞就搞出来了，最后的结果是n*(n+1)，手算一下前5个就能看出来。剩下的输出从大开始向下遍历，输出2^(len)-1-i就可以了，边计算边标记（len为当前数的2进制表示形式的位数）。

#include 
  
   
#include 
   
     #include 
    
      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<

hdu 5014 Number Sequence（意淫题）

Number Sequence