poj2406 Power Strings(KMP)

Power Strings

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.

题意：给定一个字符串，让你求出他最多由几个相同的连续子串连接而成。

分析：在这里我们假设这个字符串的长度是len，那么如果len可以被len-next[len]整除的话，我们就可以说len-next[len]就是那个最短子串的长度。

代码

#include <cstdio>
#include <string>
using namespace std;

int l,p[1000000];
char a[1000000];

int main()
{

while (scanf("%s",a),a[0]!='.')
{
l=strlen(a);
p[0]=-1;
int j=-1;
for (int i=1;i<l;i++)
{
while ((j>-1)&&(a[j+1]!=a[i])) j=p[j];
if (a[j+1]==a[i]) j++;
p[i]=j;
}
int ans=1;
if (l%(l-p[l-1]-1)==0) ans=l/(l-p[l-1]-1);
printf("%d\n",ans);
}
}