在C ++中找到最接近n且可被m整除的数字
假设我们有两个整数n和m。我们必须找到最接近n的数字并除以m。如果有多个这样的数字，则显示最大绝对值的数字。如果n可被m整除，则返回n。因此，如果n=13，m=4，则输出为12。
为了解决这个问题，我们可以按照以下步骤进行：
令q：=n/m，n1：=m*q
如果n*m>0，则n2：=m*（q+1），否则n2：=m*(q-1)
如果|n–n1|<|n–n2|，然后返回n1，否则返回n2
示例
#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;
}输出结果
Closest for n = 13, and m = 4: 12 Closest for n = 0, and m = 8: 0 Closest for n = 18, and m = -7: 21