ZOJ1905Power Strings （KMP||后缀数组+RMQ求循环节）

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.

<b< dd="">

Output

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

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

求最大循环长度。

KMP可以求，之前做过，见。

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

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

这里实现了后缀数组(不过好像被卡了，只能同KMP实现)。

#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
int min(int x,int y) { if(x<y) return x;return y;}
using namespace std;
const int maxn=;
char ch[maxn];
struct SA
{
    int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],cntA[maxn],B[maxn],cntB[maxn];
    int ht[maxn],Min[maxn][],N;
    void get_SA()
    {
        N=strlen(ch+);
        for(int i=;i<=;i++) cntA[i]=;
        for(int i=;i<=N;i++)  cntA[ch[i]]++;
        for(int i=;i<=;i++) cntA[i]+=cntA[i-];
        for(int i=N;i>=;i--)  sa[cntA[ch[i]]--]=i;
        Rank[sa[]]=;
        for(int i=;i<=N;i++)  Rank[sa[i]]=Rank[sa[i-]]+(ch[sa[i]]==ch[sa[i-]]?:);
        for(int l=;Rank[sa[N]]<N;l<<=){
            for(int i=;i<=N;i++) cntA[i]=cntB[i]=;
            for(int i=;i<=N;i++) cntA[A[i]=Rank[i]]++;
            for(int i=;i<=N;i++) cntB[B[i]=i+l<=N?Rank[i+l]:]++;
            for(int i=;i<=N;i++) cntA[i]+=cntA[i-],cntB[i]+=cntB[i-];
            for(int i=N;i>=;i--) tsa[cntB[B[i]]--]=i;
            for(int i=N;i>=;i--) sa[cntA[A[tsa[i]]]--]=tsa[i];
            Rank[sa[]]=;
            for(int i=;i<=N;i++) Rank[sa[i]]=Rank[sa[i-]]+(A[sa[i]]==A[sa[i-]]&&B[sa[i]]==B[sa[i-]]?:);
        }
    }
    void get_hgt()
    {
        for(int i=,j=;i<=N;i++){
            if(j) j--;
            while(ch[i+j]==ch[sa[Rank[i]-]+j]) j++;
            ht[Rank[i]]=j;
        }
    }
    void get_rmq()
    {
        for(int i=;i<=N;i++) Min[i][]=ht[i];
        for(int i=;(<<i)<=N;i++)
         for(int j=;j+(<<i)-<=N;j++){
              Min[j][i]=min(Min[j][i-],Min[j+(<<(i-))][i-]);
         }
    }
    int query_rmq(int L,int R)
    {
        if(L>R) swap(L,R);L++;
        int k=log2(R-L+);
        return min(Min[L][k],Min[R-(<<k)+][k]);
    }
    void solve()
    {
        int ans=;
        for(int i=;i<=N;i++){
            if(N%i!=) continue;
            if(i+query_rmq(Rank[],Rank[+i])==N) {
                ans=N/i; break;
            }
        }   printf("%d\n",ans);
    }
}Sa;
int main()
{
    while(~scanf("%s",ch+)){
        if(ch[]=='.') return ;
        Sa.get_SA();
        Sa.get_hgt();
        Sa.get_rmq();
        Sa.solve();
    }  return ;
}