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);

     }

 }