这题挺简单的

#include <bits/stdc++.h>
using namespace std;
int m, n;
int main()
{
    cin >> m >> n;
    cout << m / n << " " << m % n;
    return 0;
}

#include<bits/stdc++.h> using namespace std; int n,m; int main() { scanf("%d%d",&n,&m); cout<<n/m<<' '<<n%m; return 0; }

两种语言解法(～￣▽￣)～

1.c++代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int a , b;
    cin >> a >> b;
    cout << a / b << " " << a % b;
    return 0;
}

2.python代码

a = input().split()
b = int(a[1])
a = int(a[0])
print(a // b , a % b)

#include <iostream>//hetao3097453
using namespace std;
int main()
{
    int a,b;
    cin >> a >> b;
    cout << a / b << " " << a % b << endl;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    cin>>n>>m;
    cout<<n/m<<" "<<n-((n/m)*m);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int x,y,num,sum;
    cin>>x>>y;
    num = x%y;
    sum = x/y;
    cout << sum << " " << num;
}

#include <bits/stdc++.h>
using namespace std;
int m,n;
int main()
{
    cin>>m>>n;
    cout<<m/n<<" "<<m%n;
    return 0;
}

过辣！

    long long m,n;
    cin>>m>>n;
    cout<<m/n<<" "<<m%n;

#include <bits/stdc++.h> using namespace std; int m, n; int main() { cin >> m >> n; cout << m / n << " " << m % n; return 0; }

cout<<m/n<<" "<<m%n;