Question

我正在使用repl.it来编写我的C ++。到目前为止，我已经了解了条件，循环和函数。现在我正在尝试编写一个输入两个整数的程序，并找到最小公倍数和最大公分母。到目前为止，我已经编写了大部分代码，但是存在问题。

“以非零状态退出”

#include <iostream>

using namespace std;

int main() 
{
int number1 = 0;
int number2 = 0;
int calc = 0;
int lcm = 0;
cout << "Give me two integers, and I will calculate the Least Common Multiple and the Greatest Common Divisor." << endl;

while (number1 <= 0) {

    cout << "Enter your first number. Cant be negative" << endl;
    cin >> number1; }

while (number2 <= 0) {
    cout << "Enter your second number. Cant be negative" << endl;
    cin >> number2; }

while(number2 != 0) { ///Greatest Common Divisor
    calc = number1 % number2;
    lcm = (number1*number2) / calc;
    number1 = number2;
    number2 = calc;
}



cout << "Least Common Multiple is " << lcm << endl;

cout << "Greatest Common divisor is " << number1 << endl;
}

所以我不确定这是一个语法错误，还是因为repl.it，但我真的很难弄清楚这一点。

由于

Answer 1

while循环在循环结束时检查number2值，但是当你计算calc时，有时这个值为零，那么在下一步中程序退出时除以零异常。您可以通过在计算calc变量后将此行添加到代码来防止此问题：

if (calc == 0 ) break;

另外，您的代码无法正常工作，例如set number1 = 30和number2 = 18！

我使用二进制方法计算GCD，然后使用GCD计算LCM。

            #include <iostream>
            #include <math.h>   //  for pow(2,d)
            using namespace std;


            int main() 
            {
                int gcd, lcm, a, b, g, number1 = 0, number2 = 0, d=0;
                cout << "Give me two integers, and I will calculate the Greatest Common Divisor and the Least Common Multiple." << endl;

                while (number1 <= 0) {
                    cout << "Enter your first number. Cant be negative" << endl;
                    cin >> number1; 
                }
                // using binary method to calculating GCD:   https://en.wikipedia.org/wiki/Greatest_common_divisor
                while (number2 <= 0) {
                    cout << "Enter your second number. Cant be negative" << endl;
                    cin >> number2; 
                }    
                a = number1;
                b = number2;
                while (((a%2)==0) && ((b%2)==0)) {
                    a = a/2;
                    b = b/2;
                    d = d+1;
                }
                while (a != b) {
                    if ((a%2) == 0) {
                        a = a/2;
                    } else if ((b%2)==0) {
                        b = b/2;
                    } else if (a>b) {
                        a = (a-b) /2;
                    } else {
                        b = (b-a)/2;
                    }
                }
                g = a;
                cout << "\ng: " << g << "\td: " << d << "\tpower(2,d): " << pow(2,d);
                gcd = g * pow(2,d); // power(2,d) with math.h library
                lcm = (number1*number2)/gcd;   // according to LCM(a,b) = (a*b)/GCD(a,b)
                cout << "\nGreatest Common Divisor is " << gcd << " and Least Common Multiple is " << lcm << endl;
            }

Answer 2

首先，您需要检查哪个数字更大，因为如果您设置number1=18和number2=30，则number1%number2将为18，然后您会看到它的位置 - 到错误的输出。此外，Hossein Nazari建议的修复可以避免异常，但如果您遵循该修复，则GCD将存储在number2中。例如30和18：

30%18 = 12
calc = 12
if(calc==0) break;
(lcm routine)
number1 = 18
number2 = 12

18%12 = 6
calc=6
if(calc==0) break;
(lcm routine)
number1 = 12
number2 = 6

12%6=0
calc=0
if(calc==0) break;
<done>

我宁愿完全从你的循环中删除lcm计算，在这种情况下，你按照计划在number1得到你的GCD，然后只按公式(number1*number2)/GCD计算LCM，这将是30*18/6 = 90;

Answer 3

嗨，只需更改这两个数据类型

float calc = 0;
float lcm = 0;

嗨@Cheers和hth，我在这里执行了代码https://www.tutorialspoint.com/compile_cpp11_online.php，并得到了这个异常 - 浮点异常，在我将数据类型从int更改为float后，我得到了这个值输入您的第一个号码。不能为负 10个
输入您的第二个号码。不能为负 20个
最小公倍数是inf 最大公约数是10

C ++语法错误：程序退出非零状态

3 个答案: