Power Strings

Time Limit: 3000MS	Memory Limit: 65536K
Total Submissions: 51218	Accepted: 21386

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

題意：給出一個字串由一個子串重複n次形成，求n的最大值。

思路：列舉子串長度k，首先需要被總長度n整除，然後判斷sa[0]和sa[k]的公共字首是否n-k即可

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
using namespace std;
const int MAXN = 3e6+10;
const int N = MAXN;
int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];
char s[MAXN];
int a[MAXN];
int sa[MAXN],height[MAXN],RANK[MAXN];
int len[MAXN];
int c0(int *r,int a,int b)
{
    return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];
}
int c12(int k,int *r,int a,int b)
{
    if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);
    else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];
}
void sort(int *r,int *a,int *b,int n,int m)
{
    int i;
    for(i=0;i<n;i++) wv[i]=r[a[i]];
    for(i=0;i<m;i++) ws[i]=0;
    for(i=0;i<n;i++) ws[wv[i]]++;
    for(i=1;i<m;i++) ws[i]+=ws[i-1];
    for(i=n-1;i>=0;i--) b[--ws[wv[i]]]=a[i];
    return;
}
void dc3(int *r,int *sa,int n,int m) //涵義與DA 相同
{
    int i,j,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;
    int *rn=r+n;
    r[n]=r[n+1]=0;
    for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;
    sort(r+2,wa,wb,tbc,m);
    sort(r+1,wb,wa,tbc,m);
    sort(r,wa,wb,tbc,m);
    for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)
        rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;
    if(p<tbc) dc3(rn,san,tbc,p);
    else for(i=0;i<tbc;i++) san[rn[i]]=i;
    for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;
    if(n%3==1) wb[ta++]=n-1;
    sort(r,wb,wa,ta,m);
    for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;
    for(i=0,j=0,p=0;i<ta && j<tbc;p++)
        sa[p]=c12(wb[j]%3,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 calheight(int *r,int *sa,int n)
{   // 此處N為實際長度
    int i,j,k=0;
    // height[]的合法範圍為 1-N, 其中0是結尾加入的字元
    for(i=1;i<=n;i++)
        RANK[sa[i]]=i;
    // 根據SA求RANK
    for(i=0;i<n; height[RANK[i++]] = k )
        // 定義：h[i] = height[ RANK[i] ]
        for(k?k--:0,j=sa[RANK[i]-1];
            r[i+k]==r[j+k]; k++);
    //根據 h[i] >= h[i-1]-1 來優化計算height過程
}
int main(){
    while(scanf("%s",s)!=EOF){
        if(s[0]=='.')
            break;
        int n=(int)strlen(s);
        for(int i=0;i<n;i++){
            a[i]=s[i];
        }
        a[n]=0;
        dc3(a,sa,n+1,200);
        calheight(a, sa, n);
        int MIN=1e9+7;
        int flag=1;
        for(int i=1;i<=n;i++){
            if(sa[i]==0){
                flag=i;
                break;
            }
        }
        for(int i=flag+1;i<=n;i++){
            MIN=min(height[i],MIN);
            len[sa[i]]=MIN;
        }
        MIN=1e9+7;
        for(int i=flag-1;i>=1;i--){
            MIN=min(height[i+1],MIN);
            len[sa[i]]=MIN;
        }
        int k;
        //for(int i=0;i<=n;i++)
        //    cout<<i<<" "<<len[i]<<endl;
        for(k=1;k<=n;k++){
            if(n%k==0){
                if(len[k]==n-k){
                    break;
                }
            }
        }
        printf("%d\n",n/k);
    }
}