POJ 2406

Power Strings

Time Limit: 3000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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.

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题目大意：给定字符串，问字符串由几个相同的部分组成？

if(字符串长度%（字符串长度-字符串最后一个字符的next[]）==0),相同部分的长度为：字符串长度-字符串最后一个字符的next[]

在做这个题目时处理低级失误，把长度L定义成了char,找了一个多小时，哎……，笨的要死了！

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#include<cstdio>
#include<cstring>

using namespace std;
const int maxl=1000010;
char s[maxl];
int next[maxl],l;
void getnext()
{
    next[0]=-1;
    for(int j,i=1;i<l;i++)
    {
        j=next[i-1];
        while(s[j+1]!=s[i] && j>=0)j=next[j];
        next[i]=s[j+1]==s[i]?j+1:-1;
    }
}
int main()
{
    while(~scanf("%s",s) )
    {
        if(s[0]=='.')break;
        l=strlen(s);
        getnext();
        if(l%(l-next[l-1]-1)==0)printf("%d
",l/(l-next[l-1]-1));
        else printf("1
");
    }
    return 0;
}