A. In Search of an Easy Problem

As the name suggest it is an easy problem all what is needed is to take a string s as input and print Mohamed Salah

#include<iostream>
using namespace std;
int main()
{
 string str;cin>>str;
 cout<<"Mohamed Salah";
}

F. N and T

In this problem we want to find a number that is divisible by t and has n digits. But we found a series of problems:

1. n can be 100 -> there is no data type that handles such a big number (long long take up to pow(2,64))

2. How can I easily get a number divisible by t

The answer is t itself and

what about multiplying the number by 10 until I get the desired length? why 10: multiplying by 10 will keep the divisibility state (2x10 is divisible by 2 and 2x10x10 is still divisible by 2....)

so the answer will be the t multiplied by 10 until I reach the desired length. -> problem 2 solved

multiplying by 10 means adding 0 to the end of the number (2x10 = 20, 330x10 = 3300)

so we can print the t followed by zeros -> problem 1 solved

but when I will print -1? the answer is if the number of digits in t is less than n as the smallest number is divisble by t is t itself

#include <iostream>
#include <vector>
using namespace std ;
int32_t main() {
    int n,t;cin>>n>>t;
    if(t == 10 && n== 1){
        cout<<"-1";
    }else{
        cout<<t;
        n--;
        if(t == 10)n--;
        for (int i = 0; i < n; ++i) {
            cout<<"0";
        }
    }
}