训练题b题kmp

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.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char se[1100000];
int next[1100000];//next[i]表示前面长度为i的子串
void get(int m)
{
    int i=1,j=0;
    next[0]=0;
    while(i<m)
    {
        if(se[i]==se[j])
            next[i++]=++j;
        else if(!j)
        {
            next[i]=0;
            i++;
        }
        else
        j=next[j-1];
    }
}
int main()
{
    while(~scanf("%s",se)&&se[0]!='.')//se[0]等于.时结束
    {
        int l1,l2;//l1为前缀和后缀相等的最大长度
        l1=strlen(se);//测字符串得长度
        get(l1);
        l2=l1-next[l1-1];//最小循环节的长度 
        if(l1%l2==0)//判断一下字符串是不是由循环节组成的
            printf("%d\n",l1/l2);
        else
        printf("1\n");
    }
    return 0;
}

题意给定一个字符串m，求一个最短的子串n，m是n重复多次组成的

思路 kmp算法

kmp的next数组介绍

KMP的next数组告诉我们：当模式串中的某个字符跟主串中的某个字符失配时，模式串下一步应该跳到哪个位置。如模式串中在j 处的字符跟主串在i 处的字符匹配失配时，下一步用next [j] 处的字符继续跟主串i 处的字符匹配，相当于模式串向右移动 j - next[j] 位。

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