Zoj1905power Strings (kmp| | Suffix array +rmq to find the Loop section)

Given 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 are defined in the normal way:a^0 = "" (The empty string) and a^ (n+1) = A * (a^n).

Input

Each test case was a line of input representing S, a string of printable characters.

<b< dd= "" >

Output

For each s should print the largest n such, that s = a^n for some string a. The length of S'll be is at least 1 and would not exceed 1 million characters. A line containing a period follows the last test case.

Sample Input

Abcd
Aaaa
Ababab
.

Sample Output

1
4
3

The maximum loop length is calculated.

KMP can beg, have done before, see.

Http://www.cnblogs.com/hua-dong/p/8016873.html

Http://www.cnblogs.com/hua-dong/p/8016916.html

The suffix array is implemented here (although it seems to be stuck, it can only be implemented with KMP).

#include <cmath>#include<cstdio>#include<string>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>intMinintXintY) {if(x<y)returnXreturny;}using namespacestd;Const intmaxn=1000010;CharCH[MAXN];structsa{intRANK[MAXN],SA[MAXN],TSA[MAXN],A[MAXN],CNTA[MAXN],B[MAXN],CNTB[MAXN]; intht[maxn],min[maxn][ -],n; voidGet_sa () {N=strlen (ch+1);  for(intI=0; i<= -; i++) cnta[i]=0;  for(intI=1; i<=n;i++) cnta[ch[i]]++;  for(intI=1; i<= -; i++) cnta[i]+=cnta[i-1];  for(inti=n;i>=1; i--) sa[cnta[ch[i]]--]=i; rank[sa[1]]=1;  for(intI=2; i<=n;i++) rank[sa[i]]=rank[sa[i-1]]+ (ch[sa[i]]==ch[sa[i-1]]?0:1);  for(intL=1; rank[sa[n]]<n;l<<=1){             for(intI=1; i<=n;i++) cnta[i]=cntb[i]=0;  for(intI=1; i<=n;i++) cnta[a[i]=rank[i]]++;  for(intI=1; i<=n;i++) cntb[b[i]=i+l<=n? RANK[I+L]:0]++;  for(intI=1; i<=n;i++) cnta[i]+=cnta[i-1],cntb[i]+=cntb[i-1];  for(inti=n;i>=1; i--) tsa[cntb[b[i]]--]=i;  for(inti=n;i>=1; i--) sa[cnta[a[tsa[i]]]--]=Tsa[i]; rank[sa[1]]=1;  for(intI=2; i<=n;i++) rank[sa[i]]=rank[sa[i-1]]+ (a[sa[i]]==a[sa[i-1]]&&b[sa[i]]==b[sa[i-1]]?0:1); }    }    voidget_hgt () { for(intI=1, j=0; i<=n;i++){            if(j) j--;  while(ch[i+j]==ch[sa[rank[i]-1]+J]) J + +; Ht[rank[i]]=J; }    }    voidget_rmq () { for(intI=1; i<=n;i++) min[i][0]=Ht[i];  for(intI=1;(1<<i) <=n;i++)          for(intj=1; j+ (1<<i)-1<=n;j++) {Min[j][i]=min (min[j][i-1],min[j+ (1<< (I-1))][i-1]); }    }    intQUERY_RMQ (intLintR) {if(l>r) swap (L,R); l++; intK=LOG2 (r-l+1); returnMin (min[l][k],min[r-(1&LT;&LT;K) +1][k]); }    voidsolve () {intans=1;  for(intI=1; i<=n;i++){            if(n%i!=0)Continue; if(I+QUERY_RMQ (rank[1],rank[1+i]) = =N) {ans=n/i; Break; }} printf ("%d\n", ans); }}sa;intMain () { while(~SCANF ("%s", ch+1)){        if(ch[1]=='.')return 0;        Sa.get_sa ();        SA.GET_HGT ();        SA.GET_RMQ ();     Sa.solve (); }  return 0;}

Zoj1905power Strings (kmp| | Suffix array +rmq to find the Loop section)