[POJ

Link：http://poj.org/problem?id=2406

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 50548 Accepted: 21083

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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

题解：

KMP，next[]表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。所以思路和上面一样，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

例如：a    b    a    b    a    b

next: -1   0    0    1    2    3    4

next[n]==4,代表着，前缀abab与后缀abab相等的最长长度，这说明，ab这两个字母为一个循环节，长度=n-next[n]

Code：

#include<cstdio>#include<cstdlib>#include<algorithm>#include<iostream>#include<cmath>#include<cstring>#include<queue>#include<stack>using namespace std;const int maxn=1e6+10;char s[maxn];int next[maxn];void getnext(char *s,int m){int i=0,j=0;next[0]=-1,j=next[i];while(i<m){if(j==-1||s[i]==s[j])next[++i]=++j;elsej=next[j];}}int main(){while(~scanf("%s",s),s[0]!='.'){int l=strlen(s);getnext(s,l);int a=l-next[l];if(l%a==0)printf("%d\n",l/a);elseprintf("1\n");}return 0;}