687. Repeats spoj （后缀数组 重复次数最多的连续重复子串）

687. Repeats

Problem code: REPEATS

A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string

s = abaabaabaaba

is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.

Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string

u = babbabaabaabaabab

contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.

Input

In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.

Output

For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.

Example

Input:
1
17
b
a
b
b
a
b
a
a
b
a
a
b
a
a
b
a
b

Output:
4

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <vector>
 #include <string.h>
 using namespace std;
 #define N 50002
 char a[N];
 int c[N],d[N],e[N],sa[N],height[N],n,b[N],m,dp[N][];
 int cmp(int *r,int a,int b,int l)
 {
     return r[a]==r[b]&&r[a+l]==r[b+l];
 }
 void da()
 {
     int i,j,p,*x=c,*y=d,*t;
     memset(b,,sizeof(b));
     for(i=; i<n; i++)b[x[i]=a[i]]++;
     for(i=; i<m; i++)b[i]+=b[i-];
     for(i=n-; i>=; i--)sa[--b[x[i]]]=i;
     for(j=,p=; p<n; j*=,m=p)
     {
         for(p=,i=n-j; i<n; i++)y[p++]=i;
         for(i=; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j;
         for(i=; i<n; i++)e[i]=x[y[i]];
         for(i=; i<m; i++)b[i]=;
         for(i=; i<n; i++)b[e[i]]++;
         for(i=; i<m; i++)b[i]+=b[i-];
         for(i=n-; i>=; i--)sa[--b[e[i]]]=y[i];
         for(t=x,x=y,y=t,p=,x[sa[]]=,i=; i<n; i++)
             x[sa[i]]=cmp(y,sa[i-],sa[i],j)?p-:p++;
     }
 }
 void callheight()
 {
     int i,j,k=;
     b[]=;
     for(i=; i<n; i++)b[sa[i]]=i;
     for(i=; i<n-; height[b[i++]]=k)
         for(k?k--:,j=sa[b[i]-]; a[i+k]==a[j+k]; k++);
 }
 int fun(int i,int j)
 {
     i=b[i];
     j=b[j];
     if(i>j)swap(i,j);
     i++;
     int k=(int)(log(j-i+1.0)/log (2.0));
     return min(dp[i][k],dp[j-(<<k)+][k]);
 }
 void initrmq()
 {
     int i,j;
     memset(dp,,sizeof(dp));
     for(i=; i<=n; i++)
         dp[i][]=height[i];
     for(j=; (<<j)<=n; j++)
         for(i=; i+(<<j)<=n; i++)
             dp[i][j]=min(dp[i][j-],dp[i+(<<(j-))][j-]);
 }
 int main()
 {
     int t,i,j,r;
     scanf("%d",&t);
     for(r=; r<t; r++)
     {
         scanf("%d",&n);
         for(i=; i<n; i++)a[i]=getchar(),a[i]=getchar();
         a[n++]='\0';
         m=;
         da();
         callheight();
         initrmq();
         int max=;
         for(i=; i<n/; i++)
         {
             for(j=; j+i<n; j+=i)
             {
                 int k=fun(j,j+i);
                 int kk=k/i+;
                 int tt=i-k%i;
                 tt=j-tt;
                 if (tt>=&&k%i!=)
                     if(fun(tt,tt+i)>=k)
                         kk++;
                 if(max<kk)
                 {
                     max=kk;
                 }
             }
         }
         printf("%d\n",max);
     }
 }