POJ2406 Power Strings

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K

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

第一道kmp题。。题目意思是求一个字符串的循环节，求出循环节长度之后检验即可

#include<set>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1000010;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
char s[N];
int next[N],n;

void BuildNext(char s[]){
    next[0]=next[1]=0;
    For(i,n-1){
        int j=next[i];
        while(j&&s[i]!=s[j]) j=next[j];
        if(s[i]==s[j])  next[i+1]=j+1;
        else            next[i+1]=0;
    }
}

int main(){
    while(scanf("%s",&s),s[0]!='.'){
        n=strlen(s);BuildNext(s);
        int rpt = n-next[n];
        int i = n;
        while(i&&i-next[i]==rpt) i=next[i];
        if(i) printf("1
");
        else  printf("%d
",n/rpt);
    }
    return 0;
}