Find Closest Multiple of X to A^B in C++

Suppose we have three values, a, b and x. We have to find one multiple of x, that is nearest to a^b. Suppose the numbers are x = 4, a = 3, b = 3, then the output will be 28, as this is nearest to 3³ = 27

The approach is simple; we have to follow these conditions −

• If b < 0, and a = 1, then ab turns out to be 1 and hence, the closest multiple of x becomes either 0 or x.

• If b < 0 and a > 1, then, ab, turns out to be less than 1, and hence the closest multiple of x becomes 0.

• If b > 0, then find ab. Then let mul = integer of ab / x, then a closest multiple of x is mul*x or (mul + 1)*x

Example

 Live Demo

#include<iostream>
#include<cmath>
using namespace std;
void findMultiple(int a, int b, int x) {
   cout << "Nearest multiple: ";
   if (b < 0) {
      if (a == 1 && x == 1)
         cout << "1";
      else
         cout << "0";
   }
   int mul = pow(a, b);
   int ans = mul / x;
   int ans1 = x * ans;
   int ans2 = x * (ans + 1);
   if((mul - ans1) <= (ans2 - mul)){
      cout << ans1;
   }
   else{
      cout << ans2;
   }
}
int main() {
   int a = 3, b = 3, x = 4;
   findMultiple(a, b, x);
}

Output

Nearest multiple: 28