判断两个双精度浮点数是否整除！

判断两个双精度浮点数是否整除可以将它们相除并取余数，如果余数为0则说明它们整除。代码示例：

#include <iostream>
#include <cmath>
using namespace std;

bool isDivisible(double a, double b) {
    double quotient = a / b;
    double remainder = fmod(a, b);
    if (remainder == 0) {
        return true;
    } else {
        return false;
    }
}

int main() {
    double num1 = 12.0;
    double num2 = 3.0;
    if (isDivisible(num1, num2)) {
        cout << num1 << " is divisible by " << num2 << endl;
    } else {
        cout << num1 << " is not divisible by " << num2 << endl;
    }
    return 0;
}