【POJ 2406】Power Strings

Power Strings Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 64676 Accepted: 26679

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

题解：KMP……照着书打的qwq

#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> using namespace std; #define maxn 2000100 char s[maxn]; int nexts[maxn]; int n,m; void getnexts() { int j=0; int k=-1; nexts[0]=-1; while(j<m) { if(k==-1||s[j]==s[k]) { j++;k++; nexts[j]=k; } else k=nexts[k]; } } int main() { while(~scanf("%s",s)) { if(s[0]=='.')break; m=strlen(s); getnexts(); int n=m-nexts[m]; if(n!=m&&m%n==0) printf("%d\n",m/n); else printf("1\n"); } return 0; }

 

转载于:https://www.cnblogs.com/wuhu-JJJ/p/11149637.html

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