乙级(Basic Level) 1007

Python:
a = input().split()
a = [int(i) for i in a]
print(a[0]//a[1],a[0]%a[1])

C++:
//思路：把除法转换为求商过程的逆转,比如 100/2 等价于 1/2 商为0,余数为1,商为0的情况下不能输输出, 然后余数和下一位即0
//　　　组合为1*10+0=10,就变成10/2 商为5余数为0,此时输出商数。然后0和下一位0组合为0*10+0=0 0/2余数为0,此时运算完毕
#include<iostream>
#include<string>
using namespace std;
int main(){
    string str,ans;
    int i,j,n,d = 0;
    cin >> str >> n;
    for(i = 0; i <= str.size()-1; i++){
        int current = d * 10 + (str[i]-‘0‘);
        ans += (current / n+‘0‘);
        d = current % n;
    }
    j = 0;
    for(i=0;i<ans.size();i++)
        if(ans[i] != ‘0‘ || j){
            j = 1;
            cout<<ans[i];
        }
    cout << " " << d;
    return 0;
}