#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int x = n * n;
    int m = 2 * n + 3;
    double r = (double)m / x * 100;
    cout << fixed << setprecision(1) << r << "%";
    return 0;
}

AC code

Please give me a like

#include <bits/stdc++.h>
using namespace std;
int main()
{
    double n;
    cin >> n;
    double p = n * 2 + 3;
    cout << fixed << setprecision(1) << p / n / n * 100 << "%";
    return 0;
}

Please subscribe all of my exercise

You will be very happy like this picture

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double n,sum;
    cin>>n;
    sum=(2*n+3)*100;
    double b=sum/(n*n);
    cout<<fixed<<setprecision(1)<<b<<"%";
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int x = n * n;
    int m = 2 * n + 3;
    double r = (double)m / x * 100;
    printf("%.1f",r);
    cout << "%";
    return 0;
}

A

过辣！

    double n,s,num;
    cin>>n;
    s=n*n;
//  num=n*2-1+4;
//  可简化为 ↙
    num=n*2+3;
    cout<<fixed<<setprecision(1)<<1.0*num/s*100<<"%";

printf("%.1f",100.0/(n*n*1.0/(n*2+4-1)));
cout<<"%";