poj 2406 Power Strings (后缀数组 || KMP)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28859		Accepted: 12045

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

 //42704K    2829MS    C++    2709B    2013-12-18 13:31:43
 /*

     题意：
         给出一个由某个子串重复R次组成的字符串，求R的最大值

     后缀数组：
         KMP应该会简单些，因为此处要练习后缀数组，故用后缀数组。首先考虑用DA的话会TLE，
     因为其时间复杂度为O(nlgn)，数据太大(n=2000000)。
         此处用dc3，dc3的算法没研究，只是套用其模板，dc3算法会比DA快些，在此处勉勉强强
     的过了。先用dc3求出后缀数组，然后再求height数组，再穷举字符串长度len。求k能整除len
     且suffix(1) 与 suffix(k+1) 最长前缀等于len-k。
         在求最长公共前缀时，由于suffix(1)是固定的，利用height数组的特性，求出height数组
     中每一个数到height[rank[0]]间的最小值即可。  

 */
 #include<stdio.h>
 #include<string.h>
 #define N 2000005
 #define F(x) ((x)/3+((x)%3==1?0:tb))
 #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
 int wa[N],wb[N],wv[N],ws[N];
 int rank[N],height[N];
 int sa[N],r[N];
 char c[N];
 int lcp[N]; //记录到height[rank[0]]的最小值
 int Max(int a,int b)
 {
     return a>b?a:b;
 }
 int Min(int a,int b)
 {
     return a<b?a:b;
 }
 int cmp(int *y,int a,int b,int l)
 {
     return y[a]==y[b]&&y[a+l]==y[b+l];
 }

 int c0(int *y,int a,int b)
 {
     return y[a]==y[b]&&y[a+]==y[b+]&&y[a+]==y[b+];
 }
 int c12(int k,int *y,int a,int b)
 {
     if(k==) return y[a]<y[b]||y[a]==y[b]&&c12(,y,a+,b+);
     else return y[a]<y[b]||y[a]==y[b]&&wv[a+]<wv[b+];
 }
 void sort(int *r,int *a,int *b,int n,int m)
 {
     int i;
     for(i=;i<n;i++) wv[i]=r[a[i]];
     for(i=;i<m;i++) ws[i]=;
     for(i=;i<n;i++) ws[wv[i]]++;
     for(i=;i<m;i++) ws[i]+=ws[i-];
     for(i=n-;i>=;i--) b[--ws[wv[i]]]=a[i];
     return;
 }
 void dc3(int *r,int *sa,int n,int m)
 {
     int i,j,*rn=r+n,*san=sa+n,ta=,tb=(n+)/,tbc=,p;
     r[n]=r[n+]=;
     for(i=;i<n;i++) if(i%!=) wa[tbc++]=i;
     sort(r+,wa,wb,tbc,m);
     sort(r+,wb,wa,tbc,m);
     sort(r,wa,wb,tbc,m);
     for(p=,rn[F(wb[])]=,i=;i<tbc;i++)
         rn[F(wb[i])]=c0(r,wb[i-],wb[i])?p-:p++;
     if(p<tbc) dc3(rn,san,tbc,p);
         else for(i=;i<tbc;i++) san[rn[i]]=i;
     for(i=;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*;
     if(n%==) wb[ta++]=n-;
     sort(r,wb,wa,ta,m);
     for(i=;i<tbc;i++) wv[wb[i]=G(san[i])]=i;
     for(i=,j=,p=;i<ta && j<tbc;p++)
         sa[p]=c12(wb[j]%,r,wa[i],wb[j])?wa[i++]:wb[j++];
     for(;i<ta;p++) sa[p]=wa[i++];
     for(;j<tbc;p++) sa[p]=wb[j++];
     return;
 }
 void get_height(int n)
 {
     int i,j,k=;
     for(i=;i<=n;i++) rank[sa[i]]=i;
     for(i=;i<n;height[rank[i++]]=k)
         for(k?k--:,j=sa[rank[i]-];r[i+k]==r[j+k];k++);
     return;
 }
 int main(void)
 {
     while(scanf("%s",c)!=EOF)
     {
         if(strcmp(c,".")==) break;
         int n=strlen(c);
         for(int i=;i<n;i++)
             r[i]=c[i]+;
         r[n]=;
         dc3(r,sa,n+,);
         get_height(n);
         //for(int i=0;i<n;i++) printf("%d %d %d\n",i,rank[i],height[i]);
         memset(lcp,,sizeof(lcp));
         lcp[rank[]]=N;
         for(int i=rank[]-;i>=;i--) lcp[i]=Min(lcp[i+],height[i+]);
         for(int i=rank[]+;i<=n;i++) lcp[i]=Min(lcp[i-],height[i]);
         //for(int i=0;i<=n;i++) printf("%d %d %d\n",rank[i],height[i],lcp[i]);
         for(int k=;k<=n;k++)  //遍历所有值
             if(n%k== && lcp[rank[k]]==n-k){
                 printf("%d\n",n/k);
                 break;
             }
     }
     return ;
 }
 /*
 abcd
 aaaa
 ababab
 */