Question
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number n is almost lucky.
Input
The single line contains an integer n (1 ≤ n ≤ 1000) — the number that needs to be checked.
Output
In the only line print "YES" (without the quotes), if number n is almost lucky. Otherwise, print "NO" (without the quotes).
Example
47
YES
16
YES
78
NO
Explanation
1.Approach
Take a input n from user then then make that input n into string in different variable
Then check on each index that it have some different numbers than 4 and 7 or not if yes then make flag=1 else keep it zero only.
Then put condition if flag == 0 then cout yes
Or if flag == 1 and n% (all numbers which have 4 and 7 from 1 to 1000 put all ) in the following way
Example
if(flag==1 and (n%4==0 || n%7==0 || n%44==0|| n%47==0 || n%74==0 ||n%77==0 || n%444==0 || n%447==0 || n%474==0 || n%744==0 || n%477==0 || n%747==0 || n%774==0 || n%777==0 ))
cout<<"YES"<<endl;
else cout<<"NO"<<endl;
2.Approach
directly say ki n is divisible by all the value from 1 to 1000 which has only 4 and 7 in this way
Example
if(n%4==0 || n%7==0 || n%44==0|| n%47==0 || n%74==0 ||n%77==0 || n%444==0 || n%447==0 || n%474==0 || n%744==0 || n%477==0 || n%747==0 || n%774==0 || n%777==0 )
cout<<"YES"<<endl;
else cout<<"NO"<<endl;
Code
1.Approach Code
#include<bits/stdc++.h>
using namespace std;
#define test long long T;cin>>T;while(T--)
void solve(){
int n; cin>>n;
string x = to_string(n);
int flag=0;
for(int i=0;i<x.size();i++){
if(x[i]=='4' || x[i]=='7') continue;
else {
flag=1; break;
}
}
if(flag==1 and (n%4==0 || n%7==0 || n%44==0|| n%47==0 || n%74==0 ||n%77==0 || n%444==0 || n%447==0 || n%474==0 || n%744==0 || n%477==0 || n%747==0 || n%774==0 || n%777==0 ))
cout<<"YES"<<endl;
else if(flag==0) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
signed main() {
//test
//(if you want to take the more test cases you may uncomment it out)
solve();
}
2.Approach Code
#include<bits/stdc++.h>
using namespace std;
int main ()
{
int n ;
cin>>n;
if(n%4==0 || n%7==0 || n%44==0|| n%47==0 || n%74==0 ||n%77==0 || n%444==0 || n%447==0 || n%474==0 || n%744==0 || n%477==0 || n%747==0 || n%774==0 || n%777==0 )
{
cout<<"YES";
}
else
{
cout<<"NO";
}
return 0;
}
Hope you might liked my 1st approach which is submitted by me and it is Accepted and working 100% fine.
