poj2406解题报告

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 13765 Accepted: 5738

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

题意:求一个字符串中最多有多少重复的子串

思路：暴力搜索

#include<iostream>using namespace std;char str[1000010];int main(){int i,j,k,len;while(scanf("%s",&str[1])&&strcmp(".",&str[1])){len=strlen(&str[1]);for(i=1;i<=len;i++){if(len%i==0){for(j=1;j<=(len-i)/i;j++){for(k=1;k<=i;k++){if(str[k]!=str[j*i+k])goto end;}}cout<<len/i<<endl;goto exit;}end:i=i;}exit:i=i;}return 0;}