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Find the number closest to n and divisible by m in C++
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
Example
#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;
}Output
Closest for n = 13, and m = 4: 12 Closest for n = 0, and m = 8: 0 Closest for n = 18, and m = -7: 21
Updated on: 19-Dec-2019
1K+ Views
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