poj 2406 Power Strings(KMP变形)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28102		Accepted: 11755

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

题意：

给你一个字符串。问你这个字符串最多可以由一个子串重复多少次得到。

思路：

和大白(刘汝佳 训练指南)类似。先获取失配数组。然后匹配文本尾。len-f[i]就为循环节长度。

详细见代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=1000100;
int f[maxn];
char txt[maxn];
void getf(char *p)
{
    int i,j,m=strlen(p);
    f[0]=f[1]=0;
    for(i=1;i<m;i++)
    {
        j=f[i];
        while(j&&p[i]!=p[j])
            j=f[j];
        f[i+1]=p[i]==p[j]?j+1:0;
    }
}
int main()
{
    int len,i,ans,t;

    while(~scanf("%s",txt))
    {
        if(txt[0]=='.')
            break;
        getf(txt);
        len=strlen(txt);
        i=len;
        ans=1;
        while(f[i])
        {
            t=len-f[i];
            if(len%t==0&&len/t>ans)
               ans=len/t;
            i=f[i];
        }
        printf("%d
",ans);
    }
    return 0;
}