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Suppose we have two integers n and m. We have to find the number closest to n and divide by m. If there are more than one such number, then show the number which has maximum absolute value. If n is completely divisible by m, then return n. So if n = 13, m = 4, then output is 12.

To solve this, we can follow this steps −

- let q := n/m, and n1 := m*q
- if n * m > 0, then n2 := m * (q + 1), otherwise n2 := m * (q - 1)
- if |n – n1| < |n – n2|, then return n1, otherwise n2

#include<iostream> #include<cmath> using namespace std; int findClosest(int n, int m) { int q = n / m; int n1 = m * q; int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); if (abs(n - n1) < abs(n - n2)) return n1; return n2; } int main() { int n = 13, m = 4; cout << "Closest for n = " << n << ", and m = " << m << ": " << findClosest(n, m) << endl; n = 0; m = 8; cout << "Closest for n = " << n << ", and m = " << m << ": " << findClosest(n, m) << endl; n = 18; m = -7; cout << "Closest for n = " << n << ", and m = " << m << ": " << findClosest(n, m) << endl; }

Closest for n = 13, and m = 4: 12 Closest for n = 0, and m = 8: 0 Closest for n = 18, and m = -7: 21

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