Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 45005 | Accepted: 18792 |
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
只需要求n的错位部分n-f[n]是不是循环节
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+;
int n,f[N];
char p[N];
int main(){
int cas=;
while(true){
scanf("%s",p);
n=strlen(p);
if(p[]=='.') break;
f[]=f[]=;
for(int i=;i<n;i++){
int j=f[i];
while(j&&p[j]!=p[i]) j=f[j];
f[i+]=p[j]==p[i]?j+:;
}
if(n%(n-f[n])==) printf("%d\n",n/(n-f[n]));
else printf("1\n");
}
}