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HDOJ 5014 Number Sequence
找规律
Number Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 303 Accepted Submission(s): 149
Special Judge
Problem Description
There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:
● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
t = (a0 ⊕ b0) + (a1 ⊕ b1) +···+ (an ⊕ bn)
(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.
● ai ∈ [0,n]
● ai ≠ aj( i ≠ j )
For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):
(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 ≤ 105), The second line contains a0,a1,a2,...,an.
For each case, the first line contains an integer n(1 ≤ n ≤ 105), The second line contains a0,a1,a2,...,an.
Output
For 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
4 2 0 1 4 3
Sample Output
20 1 0 2 3 4
Source
2014 ACM/ICPC Asia Regional Xi‘an Online
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int n,a[100100],sig[100100]; long long int ans; int wei(int x) { if(x==0) return 0; return log(x*1.)/log(2.0); } int main() { while(scanf("%d",&n)!=EOF) { ans=0; memset(sig,-1,sizeof(sig)); for(int i=n;i>=0;i--) { if(n%2==0&&i==0) { sig[0]=0; continue; } if(sig[i]!=-1) { ans+=i^sig[i]; continue; } int w=wei(i); w++; int fan=((1<<w)-1)^i; sig[i]=fan; sig[fan]=i; ans+=i^sig[i]; } printf("%I64d\n",ans); for(int i=0;i<=n;i++) { int x; scanf("%d",&x); if(i) putchar(32); printf("%d",sig[x]); } putchar(10); } return 0; }
HDOJ 5014 Number Sequence
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