Maximum repetition substring (poj3693 后缀数组求重复次数最多的连续重复子串）

Maximum repetition substring

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6328		Accepted: 1912

Description

The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

Input

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a '#'.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

Sample Input

ccabababc
daabbccaa
#

Sample Output

Case 1: ababab
Case 2: aa

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <vector>
 #include <string.h>
 using namespace std;
 #define N 100005
 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,yy;
     while(~scanf("%s",a))
     {
         yy=;
         if(a[]=='#')break;
         n=strlen(a);
         n++;
         m=;
         da();
         callheight();
         initrmq();
         memset(c,,sizeof(c));
         int max=;
         n--;
         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)
                 {
                     yy=;
                     c[yy++]=i;
                     max=kk;
                 }
                 else if(max==kk)
                 {
                     c[yy++]=i;
                 }
             }
         }
         printf("Case %d: ",t++);
         int sta,m;
         for(i=; i<n; i++)
         {
             for(j=; j<yy; j++)
             {

                 if(fun(sa[i],sa[i]+c[j])>=(max-)*c[j])
                 {
                     sta=sa[i];
                     m=max*c[j];
                     break;
                 }
             }
             if(j<yy)break;
         }
         for(i=; i<m; i++)putchar(a[sta+i]);
         printf("\n");
     }
 }