poj 2406 Power Strings(KMP变形)

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal 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

abcdaaaaababab.

Sample Output

143

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\n",ans);    }    return 0;}