#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    if(n==1)cout<<5<<".0";
    else if(n>=2&&n<=5)printf("%.1f",4.5*n);
    else if(n>=6&&n<=10)cout<<4*n<<".0";
    else printf("%.1f",3.5*n);
    return 0;
}

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#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n == 1)
    {
        cout << "5.0";
    }
    if(n >= 2 and n <= 5)
    {
        cout << fixed << setprecision(1) << n * 4.5;
    }
    if(n >= 6 and n <= 10)
    {
        cout << fixed << setprecision(1) << n * 4.0;
    }
    if(n > 10)
    {
        cout << fixed << setprecision(1) << n * 3.5;
    }
    return 0;
}//已AC

#include <iostream>//hetao3097453
#include <iomanip>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n == 1)
    {
        cout << fixed << setprecision(1) << 1 * 5.0;
    }
    else if(n == 2 || n == 3 || n == 4 || n == 5)
    {
        cout << fixed << setprecision(1) << n * 4.5;
    }
    else if(n == 6 || n == 7 || n == 8 || n == 9 || n == 10)
    {
        cout << fixed << setprecision(1) << n * 4.0;
    }
    else
    {
        cout << fixed << setprecision(1) << n * 3.5;
    }
    return 0;
}

m=input() a=int(m) if a1: print(5.0) elif a>1 and a<6: a = 4.5 print(a) elif a>5 and a<10: print(a4.0) else: print(a*3.5) '''**注释： 因为本题要求保留一位小数， 在Python中，浮点数（一位)*整数仍得一位浮点数，所以可以把输入的瓶数乘以一位浮点数（如a9，a就乘以4.0），得到的就是一位浮点数，满足题意。

'''

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double x;
    cin>>x;
    if (x==1)
        cout<<fixed<<setprecision(1)<<5.0;
    else if (x<=5)
        cout<<fixed<<setprecision(1)<<4.5*x;
    else if (x<=10)
        cout<<fixed<<setprecision(1)<<4*x;
    else
        cout<<fixed<<setprecision(1)<<3.5*x;
    return 0;
}