`` 最基础的数学题

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double n;
    cin >> n;
    double t = n / 3.14 / 2;
    cout << fixed << setprecision(2) << t * t * 3.14;
    return 0;
}

很简单

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    double a,b,d;
    cin>>a;
    b=a/3.14/2;
    d=3.14*b*b;
    printf("%.2f",d);
    return 0;
}

上代码~

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    double x,n,sum;
    cin>>x;
    n = x/2/3.14;
    sum = n*n*3.14;
    printf("%.2f",float(sum));
    return 0;
}

已AC，放心食用~

世界最短代码！

#include <bits/stdc++.h>
using namespace std;
int main(){
    cout<<fixed<<setprecision(2)<<314.00;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double n;
    cin>>n;
    double r=(n/3.14)/2;
    double s=r*r*3.14;
    cout<<fixed<<setprecision(2)<<s;
    return 0;
}

过辣！

    double c,r;
    cin>>c;
    r=1.0*c/6.28;
    cout<<fixed<<setprecision(2)<<1.0*3.14*pow(r,2);//pow函数求r的2次方

不是很难 ···#include<bits/stdc++.h> using namespace std; int main() { double a; cin >> a; double t = a / 3.14 / 2; cout << fixed << setprecision(2) << t * t * 3.14; return 0; } ···

#include <bits/stdc++.h> using namespace std; int main() { double x; cin>>x; double ban=x/6.28; double s=3.14banban; cout << fixed << setprecision(2) << s; return 0;

}

号没了

#include <bits/stdc++.h>
using namespace std;
int main()
{
    double n,a,b;
    cin >> n;
    a=n/3.14;
    b=a/2;
    cout << fixed << setprecision(2);
    cout << b*b*3.14;

printf("%.2f",(x/2)*(x/2)/3.14);