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<
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