poj 2406 Power Strings（KMP入门，next函数理解）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37685		Accepted: 15590

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

基础题目，只要是理解next函数。

 #include<iostream>
 #include<vector>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <math.h>
 #include<algorithm>
 #define ll long long
 #define eps 1e-8
 using namespace std;

 int nexts[];
 char b[];

 void pre_nexts(int m)
 {
     memset(nexts,,sizeof(nexts));
     int j = ,k = -;
     nexts[] = -;
     while(j < m)
     {
         if(k == - || b[j] == b[k])  nexts[++j] = ++k;
         else k = nexts[k];
     }
 }
 int main(void)
 {
     while(scanf("%s",b),b[] != '.')
     {
         int n = (int)strlen(b);
         pre_nexts(n);
         if(n % (n-nexts[n]) ==  && n/(n - nexts[n]) > ) printf("%d\n",n/(n-nexts[n]));
         //根据next函数的最后一个匹配数，算出循环节点～
         else printf("1\n");
     }
     return ;
 }