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Find N digits number which is divisible by D in C++
Suppose we have two numbers N and D. We have to find N digit number, that is divisible by D. If N is 3, and D is 5, then the number can be 500. This can be solved easily. If D is 10 and N is 1, then it will be impossible. We can put D, and suppose the D has m number of digits, then attach N – m number of 0s to make it N digit number and divisible by D.
Example
#include<iostream>
using namespace std;
string nDigitDivByD(int n, int d) {
string ans = "";
if (d < 10) {
ans += to_string(d);
for (int i = 1; i < n; i++)
ans += "0";
}
else {
if (n == 1)
return "Cannot find any number";
else {
string temp = to_string(d);
ans += to_string(d);
for (int i = 0; i < n-temp.length(); i++)
ans += "0";
}
}
return ans;
}
int main() {
int n = 5, d = 15;
cout << nDigitDivByD(n, d);
}Output
15000
Updated on: 01-Nov-2019
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