POJ 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.

题目大意

给你任意个长度<=1000000的字符串，求每个字符串有多少个循环节。输入以输入一个“.”结束。

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.

题解

用kmp求next数组，通过该数组可以看出该字符串中的循环节。若字符串长度l不能整除循环节长度m，则输出1，否则输出l/m。

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;char a[1000005];int next[1000005],l;void prep(){next[1]=0; int j=0;for(int i=2;i<=l;i++)   {while(j>0&&a[i]!=a[j+1]) j=next[j];    if(a[i]==a[j+1]) j++;    next[i]=j;   }}int main(){while(scanf("%s",a+1)!=EOF)   {if(a[1]=='.') break;    l=strlen(a+1);    prep();    if(l%(l-next[l])==0) printf("%d\n",l/(l-next[l]));    else printf("1\n");   }return 0;}