Power Strings
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 58112 | Accepted: 24140 |
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
#include<stdio.h>
#include<string.h>
#include<iostream>
typedef long long ll;
using namespace std;
const int MAXN=1000000;
char str[MAXN];
int next[MAXN];
int len;
void getNext(){
int j,k;
j=0;
k=-1;
next[0]=-1;
while(j<len){
if(k==-1||str[j]==str[k])
next[++j]=++k;
else k=next[k];
}
}
int main(){
while(scanf("%s",&str)!=EOF){
if(strcmp(str,".")==0) break;
len=strlen(str);
getNext();
if(len%(len-next[len])==0&&len/(len-next[len])>1)printf("%d\n",len/(len-next[len]));
else printf("1\n");
}
return 0;
}