题解 | #时间转换#

#include<math.h>
using namespace std;
int main()
{
    unsigned long a, sum = 0,  arr[20] = { 0 };
    int i;
    cin >> a ;
    for (i = 0; i < 20; i++)
    {
        if (a != 0)
        {
            arr[i] = a % 2;
            a /= 10;
        }
        else 
            break;
    }
    for (i--; i >= 0; i--)
    {
        sum += arr[i]*pow(10,i);
    }
    cout << sum << endl;
    system("pause");
    return 0;
}


//            2.通过递归函数实现
#include<iostream>
using namespace std;
unsigned long set(unsigned long a, unsigned long sum = 0)
{
    if (a != 0)
    {
        sum = a % 2;
        sum += set(a / 10) * 10;
    }
    return sum;
}

int main()
{
    unsigned long a, sum;
    cin >> a;
    sum = set(a);
    cout << sum << endl;
    system("pause");
    return 0;
}