闰年练习。但是 "else" 必须打印下一个闰年还剩多少年

Question

所以我几乎才刚刚开始编码。我想知道你们中是否有人可以帮助我处理这个简单的代码。我需要说明离下一个闰年还有多少年，我迷路了。

int main()
{

int year;

cout << "Enter a year: ";
cin >> year;

if (year % 4 == 0)
{
    if (year % 100 == 0)
    {
        if (year % 400 == 0)
            cout << year << " is a leap year.";
        else
            cout << "There is " << 4%-year << " years till next leap year";
    }
    else
        cout << year << " is a leap year.";
}
else
   cout << "There is " << ???year << " years till next leap year";

return 0;

}

Answer 1

我已经修改了你的代码。请使用以下代码并尝试理解您的错误：-

int main()
{

    int year;

    cout << "Enter a year: ";
    cin >> year;
    int isleap = year % 4;
    if (isleap == 0)
    {
        isleap = year % 100;
        if (isleap == 0)
        {
            isleap = year % 400;
            if (isleap == 0)
                cout << year << " is a leap year."<<endl;
            else
                cout << "There is " << (4-isleap) << " years till next leap year"<<endl;
        }
        else
            cout << year << " is a leap year."<<endl;
    }
    else
       cout << "There is " << (4-isleap) << " years till next leap year"<<endl;

    return 0;

}

闰年练习。但是 "else" 必须打印下一个闰年还剩多少年

Leap year exercise. but "else" has to print how many years there is left to next leap year

c++

leap-year