垃圾代码，不喜勿喷

非常简单好吧~

#include<bits/stdc++.h> using namespace std; int main(){ double r,pl = 3.14159; scanf("%lf",&r); printf("%.4lf",r * 2); cout << " "; printf("%.4lf",r * 2 * pl); cout << " "; printf("%.4lf",r * r * pl); return 0; }

题解 用高精度

#include <bits/stdc++.h> using namespace std; int main(){ double r; cin>>r; printf("%.4f",r2);//直径 cout<<" "; printf("%.4f",23.14159r);//周长 cout<<" "; printf("%.4f",rr*3.14159);//面积 return 0; }

#include<iostream> #include<cstdio> using namespace std; int main(){ double r,d,c,s,pi=3.14159; cin>>r; d=r2; c=r2pi; s=rr*pi; printf("%.4f %.4f %.4f",d,c,s); return 0; }

#include<iostream> #include<cstdio> using namespace std; int main(){ double r, d = 0, c = 0, s = 0; double PI = 3.14159; cin >> r; d = r * 2; c = r * 2 *PI; s = r * r * PI; printf("%.4f", d); cout << " "; printf("%.4f", c); cout << " "; printf("%.4f", s); return 0; }

r=input() r=float(r) d='%.4f' % (r*2) c=float(d)*3.14159 c='%.4f' % c s=(r**2)*3.14159 s='%.4f' % s print(d,c,s)

#include<iostream> using namespace std; int main(){ double a; cin>>a; double b=a2; double PI=3.14159; double c=a2PI; double d=aa*PI; printf("%.4lf",b); cout<<" "; printf("%.4lf",c); cout<<" "; printf("%.4lf",d); return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ double r,pl = 3.14159; scanf("%lf",&r); printf("%.4lf",r * 2); cout << " "; printf("%.4lf",r * 2 * pl); cout << " "; printf("%.4lf",r * r * pl); return 0; }

#include<iostream> using namespace std; int main(){ double a,b,c,d; cin>>a; b=a2; c=2a3.14159; d=aa*3.14159; printf("%.4f",b); cout<<" "; printf("%.4f",c); cout<<" "; printf("%.4f",d); return 0; }

r = float(input()) d = "{:.4f}".format(2r) C = "{:.4f}".format(2r3.14159) S = "{:.4f}".format(3.14159r*r) print(d + " " + C + " " +S)