c++ - 第一个数和第二个数之间可同时被 4 和 6 整除的所有整数之和

Question

我需要编写一个程序来获取 2 个整数。然后程序将显示两个数字之间可同时被 4 和 6 整除的所有整数的总和

我尝试执行下面的代码:

#include <iostream>
using namespace std;

int main()
{
    int num1, num2, sum=0;

    cout << "Input first number : ";
    cin >> num1;
    cout << "Input second number : ";
    cin >> num2;


    for(int i = num1 + 1;i<num2;i++)
    {
        if ( num1 % 4 == 0 && num2 % 6 == 0)
        {
            sum = sum + i;
        }
    }
    cout<< "The sum of all integers that are both divisible by 4 and 6 between " << num1 << " and " << num2 << " is " << sum << endl;
    system("pause");
    return 0;
}

我的预期结果应该是

输入第一个数字:4
输入第二个数字:12
4和12之间所有能被4和6整除的整数之和是12“因为 12 是唯一可以同时被 4 和 6 整除的数字”

但实际结果是
输入第一个数字:4
输入第二个数字:12
4和12之间所有能被4和6整除的整数之和是56

Answer 1

这是正确的程序:

#include <iostream>
using namespace std;

int main()
{
    int num1, num2, sum=0;

    cout << "Input first number : ";
    cin >> num1;
    cout << "Input second number : ";
    cin >> num2;


    for(int i = num1;i<=num2;i++) // if you want to include num1 and num2
    //for(int i = num1+1;i<num2;i++) // if you do not want to include num1 and num2
    {
        if ( i % 4 == 0 && i % 6 == 0)
        {
            sum = sum + i;

        }
    }
    cout<< "The sum of all integers that are both divisible by 4 and 6 between " << num1 << " and " << num2 << " is " << sum << endl;
    system("pause");
    return 0;
}