#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<string>
#include<cstring>


#define PR(x) printf("")
#define READ2(x,y) scanf("%d %d",&x,&y)
#define REP(i,a) for(int i=0;i<a;i++)
#define READ(x) scanf("%d",&x)
#define PRARR(x,n) for(int i=0;i<n;i++)printf("")
using namespace std;
typedef long long ll;
int main() { 
string s1,s2;
cin>>s1;cin>>s2;
int idx=1;
string subst;
char ch=s1[0];
int len1=s1.length();
int len2=s2.length();
int lenC=1;
while(idx<len1&&ch!=s1[idx++]) lenC++;
PR(lenC);
if(lenC!=len1) {

for(int j=1;j<lenC&&j<len1;j++){
while(len1%lenC!=0) lenC++;
for(int k=j+lenC;k<len1;k+=lenC){
if(s1[j]!=s1[k]) {
//printf("VIolation at %d %d char values %c %c\n",j,k,s1[j],s1[k]);
k-=lenC;
lenC+=1;

}

}
}

}
lenC=min(len1,lenC);
int count1=0;
for(int i=1;i*lenC<=len1;i++) count1++;
PR(lenC);
PR(count1);
subst=s1.substr(0,lenC);
if(len2%lenC!=0||len2<lenC) {
			puts("0");
			return 0;
		}
bool flg=false;
for(int i=0;i<lenC;i++){
for(int j=i;j<len2;j+=lenC)
if(s2[j]!=subst[i]){
//printf("VIolation at %d %d char values %c %c\n",i,j,s2[j],subst[i]);
flg=true;
break;
}
if(flg) break;
}
if(flg) {
		puts("0");
		return 0;
	}
int num2=0;
for(int i=1;i*lenC<=len2||i*lenC<=len1;i++) if(len2%(i*lenC)==0&&len1%(i*lenC)==0)num2++;
int ans=num2;
printf("%d\n",ans);
}

I am getting wrong answer on pretest 9.The pretest is very long so can't see the complete test case. My solution is based on the following approach 1.if a substring x is a divisor of a string S then it will be either a.Shortest divisor of S(in terms of length) b.It can be broken down into smaller divisors based on the constraint for all divisor length l S.length()%l==0

I find out the smallest divisor in either string,then check if it is a divisor in the other string,if it is not then the strings dont have any common divisor.In case it is a divisor then i combine the smaller divisor to get bigger divisors such that the length of bigger divisors divide both S1.length() and S2.length() where S1 and S2 are the input strings.