【POJ2406】 Power Strings （KMP）

Power Strings

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.

 

 

【题意】

　　一个字符串的最大重复字串。

　　即一个字符串是一个子串最多重复多少次得到的。

 

【分析】

　　

　　① len==1||s[next[len-1]]!=s[len-1] ans=1；
　　② len%(len-next[len])==0 ans= len/(len-next[len]);

　　原因如图：

　　

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define Maxn 1000010

char s[Maxn];
int len,nt[Maxn];


void kmp()
{
    nt[1]=0;
    int p=0;
    for(int i=2;i<=len;i++)
    {
        while(s[i]!=s[p+1]&&p) p=nt[p];
        if(s[i]==s[p+1]) p++;
        nt[i]=p;
    }
}

int main()
{
    int n,i;
    while(1)
    {
        gets(s+1);
        len=strlen(s+1);
        if(len==1&&s[1]=='.') break;
        kmp();
        if(len%(len-nt[len])==0) printf("%d
",len/(len-nt[len]));
        else printf("1
");
    }
}

2016-07-20 16:30:02