poj--2406

Power Strings

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 38544 Accepted: 16001

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

题目来源:http://poj.org/problem?id=2406

考查点:Kmp中next数组的应用

参考资料：http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html

代码如下：

#include<stdio.h>
#include<string.h>
char a[1000010];
int per[1000010];
int b[1000010];
int l;
void getp()
{
int i,j;
j=-1;i=0;
per[i]=-1;
while(i<l)
{
if(j==-1||a[i]==a[j])
{
i++;
j++;
per[i]=j;
}
else j=per[j];
}
}
int main()
{
while(scanf("%s",a))
{
if(strcmp(a,".")==0)break;
l=strlen(a);
memset(per,0,sizeof(per));
getp();
if(l%(l-per[l])==0)
printf("%d\n",l/(l-per[l]));
else printf("1\n");
}
return 0;
}