poj 2406 Power Strings 【kmp】

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37564		Accepted: 15532

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

分析：
查找最大序列数。
代码：

#include<cstdio>
#include<cstring>
const int max= 1000100;
char str[max];
int p[max];
int cnt;
int len;
void getp()
{
len=strlen(str);
int i=0,j=-1;
p[0]=-1;
while(i<len)
{
if(j==-1||str[i]==str[j])
{
i++,j++;
p[i]=j;
}
else j=p[j];
}
}
int main()
{
while(scanf("%s",str)!=EOF)
{
if(strcmp(str,".")==0)
break;
getp();
cnt=1;
//		for(int i=0;i<len;i++)
//		printf("%d ",p[i]);
if(len%(len-p[len])==0)
cnt=len/(len-p[len]);
printf("%d\n",cnt);
}
return 0;
}