POJ 2406 kmp简单应用

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29067		Accepted: 12143

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

已知一个字符串由多个子串重复若干次得到，求最大的重复次数，

思路：可以用kmp，也可以用后缀数组，后缀数组以后实现，kmp根据next数组可以得到循环节，进而可以得到重复次数。

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014-1-29 18:06:17
File Name :4.cpp
************************************************ */

#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxn=1002000;
char str[maxn];
int next[maxn];
int get_next(char *str)
{
int j=0,k=-1;
int len=strlen(str);
next[0]=-1;
while(j<len)
{
if(k==-1||str[j]==str[k])
next[++j]=++k;
else k=next[k];
}
j=len-k;
if(len%j==0)
return len/j;
return 1;
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
while(~scanf("%s",str)&&str[0]!='.')printf("%d\n",get_next(str));
return 0;
}