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Hotaru's problem
Hotaru‘s problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3314 Accepted Submission(s): 1101
Problem Description
Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let‘s define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Let‘s define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Input
There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases.
For each test case:
the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.
For each test case:
the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.
Output
Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.
We guarantee that the sum of all answers is less than 800000.
We guarantee that the sum of all answers is less than 800000.
Sample Input
1102 3 4 4 3 2 2 3 4 4
Sample Output
Case #1: 9
Author
UESTC
Source
2015 Multi-University Training Contest 7
//在manachar的基础上，枚举回文串的中心，再找第三部分。 #include<cstdio>#include<cstring>#include<iostream>using namespace std;int read(){ register int x=0;bool f=1; register char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return f?x:-x;}const int N=3e5+10;int n,ans,cas,l,T,s[N],S[N],p[N];void manacher(){ int id=0,mx=-1; for(int i=1;i<l;i++){ if(id+mx>i) p[i]=min(p[id*2-i],id+mx-i); while(i-p[i]>=0&&i+p[i]<=l&&S[i-p[i]]==S[i+p[i]]) p[i]++; if(id+mx<i+p[i]) id=i,mx=p[i]; }}void init(){ l=0;memset(p,0,sizeof p); for(int i=0;i<n;i++) S[++l]=-1,S[++l]=s[i]; S[++l]=-1; }int main(){ for(T=read(),cas=1;ans=0,cas<=T;cas++){ n=read(); for(int i=0;i<n;i++) s[i]=read(); init();manacher(); for(int i=1;i<=n*2+1;i+=2){ for(int j=i+p[i]-1;j-i>ans;j-=2){ if(j-i+1<=p[j]){ ans=max(ans,j-i); break; } } } printf("Case #%d: %d\n",cas,ans/2*3); } return 0;}
Hotaru's problem
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