//#include<iostream> #include<cstdio> //#include<algorithm> //#include<cmath> using namespace std; const double Pi=3.14159; int main() { double r; scanf("%lf",&r); printf("%.4lf %.4lf %.4lf",2r,2rPi,Pir*r); return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ double a; cin>>a; printf("%.4f",a2); cout<<" "; printf("%.4f",a23.14159); cout<<" "; printf("%.4f",aa*3.14159); return 0; }

#include <iostream> #include <iomanip> using namespace std; int main() { double PI,r; cin>>r; PI=3.14159; cout<<fixed<<setprecision(4)<<r2; cout<<' '; cout<<fixed<<setprecision(4)<<r2PI; cout<<' '; cout<<fixed<<setprecision(4)<<rr*PI; return 0; }

不需要太多头文件 (也可以直接用万能头 # include <bits/stdc++.h> 我这里就不用万能头了 （已缩进，可抄） ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

#include<iostream> #include <iomanip> using namespace std; int main(){ double a,pi=3.14159; cin >> a; cout << fixed<<setprecision(4)<<2a << " " <<fixed<<setprecision(4)<< 2api << " " << fixed<<setprecision(4)<<aa*pi; return 0; }

先分别吧直径、周长、面积算出来再用printf来保留小数 代码如下： 看了我题解后： 错误人数 正确人数 0 114514

#include <bits/stdc++.h> using namespace std; int main(){ double r; cin>>r; printf("%.4f %.4f %.4f",r2,r23.14159,rr*3.14159); 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; c=a2; b=a23.14159; d=aa*3.14159; printf("%.4f ", c ); printf("%.4f ", b ); printf("%.4f ", 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<bits/stdc++.h> using namespace std; int main(){ double r; ？？？？？("%？？",&r); double a=2.0*？; double b=2.0r3.14159; double c=rr3.14159; printf("%.？？ %.4f %.4f",a,b,c); return 0; }

反手一个命名空间

#include<bits/stdc++.h> using namespace std; int main() { double r,pi=3.14159,zj,zc,mj; cin>>r; zj=r2; zc=r2pi; mj=rr*pi; printf("%.4f",zj); cout<<" "; printf("%.4f",zc); cout<<" "; printf("%.4f",mj); cout<<" "; }

不用万能头