横着跑一遍 , 求出每个的最小公倍数 , 再竖着跑一遍
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#define M 80
#define N 10050
using namespace std;
string s[N] ;int n,m,lcm1=1,lcm2=1,nxt[N];
int gcd(int a,int b){return !b?a:gcd(b,a%b);}
int Lcm(int a,int b){return a/gcd(a,b)*b;}
int KMP(string s){
memset(nxt,0,sizeof(nxt));
for(int i=1,j=0;i<m;i++){
while(j && s[i]!=s[j]) j = nxt[j-1];
if(s[i]==s[j]) j++;
nxt[i] = j;
}return m - nxt[m-1];
}
int KMP2(int k){
memset(nxt,0,sizeof(nxt));
for(int i=1,j=0;i<n;i++){
while(j && s[i][k]!=s[j][k]) j = nxt[j-1];
if(s[i][k]==s[j][k]) j++;
nxt[i] = j;
}return n - nxt[n-1];
}
int main(){
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
cin>>s[i];
lcm1 = Lcm (lcm1 , KMP(s[i]));
if(lcm1>=m){lcm1=m; break;}
}
for(int i=0;i<m;i++){
lcm2 = Lcm (lcm2 , KMP2(i));
if(lcm2>=n){lcm2=n; break;}
}
printf("%d",lcm1*lcm2); return 0;
}