Algorithm
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.
Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.
The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.
Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
3 2
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Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, t;
cin>>n>>t;
if(n == 1 && t == 10){
cout<<-1<<endl;
return 0;
}
else if(n == 1){
cout<<t<<endl;
return 0;
}
if(t == 10){
t = 1;
}
cout<<t;
for(int i = 0; i < n-1; i++){
cout<<0;
}
}Input
Output
Demonstration
Codeforcess Solution 584-A A. Olesya and Rodion ,C++, Java, Js and Python ,584-A,Codeforcess Solution