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Add N digits to A such that it is divisible by B after each addition?
Given a, b and n. And we have to consider the following conditions and find the optimal solution to add n digits to a such that it is divisible by b after every iteration.
Add a digit to a in such a way that after adding it, a is divisible by b.
Print the smallest value of a possible after n iterations of step1.
Print fail if the operation fails.
check divisibility after every digit addition.
Input
a=5 b=4 n=4
Output
52000
Explanation
The first digit to be added from 0 to 9, if none of the digits make a divisible by b then the answer is -1 which means the if n digits are added in a. a never be divided by b . else add the first digit that satisfies the condition and then add 0 after that (n-1) times because if a is divisible by b then a*10, a*100, … will also be divisible by b.
Example
#include <iostream>
using namespace std;
int main() {
int a = 5, b = 4, n = 4;
int num = a;
for (int i = 0; i <= 9; i++) {
int temp = a * 10 + i;
if (temp % b == 0) {
a = temp;
break;
}
}
if (num == a) {
a = -1;
}
for (int j = 0; j < n - 1; j++) {
a *= 10;
}
if(a>-1) {
cout <<a;
} else {
cout <<”fail”;
}
return 0;
}
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