思路

$C=2\pi r,S=\pi r^2$。

code

#include<bits/stdc++.h>
using namespace std;
int r;
const double p=3.1415926;
int main(){
	cin>>r;
	printf("%.2lf %.2lf",2*p*r,r*r*p);
	return 0;
}

#include<bits/stdc++.h> using namespace std; int main() { double r,c,s; const double pi=3.1415926; cin>>r; c=2pir; s=pirr; cout<<fixed<<setprecision(2)<<c<<" "<<s<<endl; return 0; }

#include<bits/stdc++.h> using namespace std; const double pi=3.1415926; int main() { double r,c,s; cin>>r; c=2rpi; s=pirr; cout<<fixed<<setprecision(2)<<c<<" "; cout<<fixed<<setprecision(2)<<s; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { double r,c,s,PI; cin>>r;PI=3.1415926; c=2rPI ;s=PIrr; cout<<fixed<<setprecision(2)<<c<<" "<<s; return 0; }

NEW right

#include<bits/stdc++.h>
using namespace std;
 int main() 
{ double r,a,b,c;
 cin>>r; 
a=3.1415926;
 b=2*a*r;
 c=a*r*r; 
cout<<fixed<<setprecision(2)<<b<<" "<<c;
 return 0;
 }

#include <stdio.h>#include <stdlib.h>int main()float r,p=3.14159,C,S;printf("输入圆的半径r\n"；scanf("%f",&r);C=2pr;S=prr;printf("圆的周长C=%f\n"，C);printf("圆的面积S=%f\n",S)；return 0;

#include<iostream> #include<iomanip> using namespace std; int main() { double r,a,b,c; cin>>r; a=3.1415926; b=2ar; c=arr; cout<<fixed<<setprecision(2)<<b<<" "<<c; return 0; }