POJ - 2406 Power Strings

Power Strings

Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

SubmitStatus

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

abcdaaaaababab.

Sample Output

143考察kmp中nex数组的性质1,nex[a]=b代表a之前的一段字符和b之前的一段字符相等。2,如果nex[len-1]=l-2;说明字符串是个形如aaaaa，bbbbb的串3,如果有循环节，那么nex[len-1]=b时，循环节k为len-1-nex[len-1];若len%k=0,说明循环，否则不循环#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int nex[1009904];char s[1000904];int l;int a,b,c,d,n,m;void getnex(){        int j=-1;        for (int i=0;i<l;i++)                {                while(j!=-1 && s[i]!=s[j+1])                        j=nex[j];                if (s[i]==s[j+1] && i!=0)                        j++;                nex[i]=j;                }}int main(){        while(1)                {                nex[0]=-1;                scanf("%s",&s);                if (s[0]=='.')                        return 0;                l=strlen(s);                getnex();                int a;                a=nex[l-1];                a=l-a-1;                if (l%a==0)     cout<<l/a<<endl;                else cout<<1<<endl;                }}