Total Submission(s) : 10 Accepted Submission(s) : 4
Problem 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
#include<iostream> #include<string.h> using namespace std; int next[1000005],len; void set_naxt(char str[]) { int i=0,j=-1; next[0]=-1; len=strlen(str); while(i<len) { if(j==-1||str[i]==str[j]) { i++; j++; next[i]=j; } else j=next[j]; } } int main() { int I; char str[1000001]; while(cin>>str&&strcmp(str,".")!=0) { set_naxt(str); if(len%(len-next[len])==0) I=len/(len-next[len]); else I=1; cout<<I<<endl; } } /*len=strlen(str); for(i=1;i<=len;i++) { for(j=i;j<len;j++) if(str[j]!=str[j%i]) break; if(j==len) break; } */ #include<iostream>
#include<string.h>
using namespace std;
int next[1000005],len;
void set_naxt(char str[])
{
int i=0,j=-1;
next[0]=-1;
len=strlen(str);
while(i<len)
{
if(j==-1||str[i]==str[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int I;
char str[1000001];
while(cin>>str&&strcmp(str,".")!=0)
{
set_naxt(str);
if(len%(len-next[len])==0)
I=len/(len-next[len]);
else
I=1;
cout<<I<<endl;
}
}
/*len=strlen(str);
for(i=1;i<=len;i++)
{
for(j=i;j<len;j++)
if(str[j]!=str[j%i])
break;
if(j==len)
break;
}
*/
