#include <bits/stdc++.h>
using namespace std;
int main()
{
    int x;
    cin >> x;
    if (x % 10 == 0)
    {
        cout << x * 0.8;
    }
    else
    {
        cout << x;
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if (n%10==0)
    {
        cout<<n*0.8;
    }
    else
    {
        cout<<n;
    }
    return 0;
}//AC

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n % 10 == 0)
    {
        cout << n * 0.8;
    }
    else
    {
        cout << n;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int x;
    cin>>x;
    if(x % 10 == 0)
    {
        cout << x * 0.8;
    }
    else
    {
        cout << x;
    }
    return 0;
}
//非常简单

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int c;
    cin >> c;
    if(c % 10 == 0)
    {
        c *= 0.8;
        cout <<c;
    }
    else
    {
        cout << c;
    }
    return 0;
}

A

#include <bits/stdc++.h>
using namespace std;
int n;
int main()
{
    cin >> n;
    if (n % 10 == 0)
    {
        cout << 0.8 * n;
    }
    else
    {
        cout << n;
    }
    return 0;
}

然而我发现这里好像不用用double()

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int x;
    cin>>x;
    double y=x;
    if (x%10==0)
        y*=0.8;
    cout<<y;
    return 0;
}