AcWing 613. 面积 (Java版和C++版) 题解
原题链接
简单
作者:
林小鹿
,
2020-10-24 19:57:59
,
所有人可见
,
阅读 461
Java 代码
import java.util.Scanner;
import java.text.DecimalFormat;
public class Main
{
public static void main(String[] args)
{
final double Pi=3.14159;
Scanner cin=new Scanner(System.in);
double a,b,c;
a=cin.nextDouble();
b=cin.nextDouble();
c=cin.nextDouble();
DecimalFormat df=new DecimalFormat("0.000");
System.out.println("TRIANGULO: "+df.format(a*c/2));
System.out.println("CIRCULO: "+df.format(Pi*c*c));
System.out.println("TRAPEZIO: "+df.format((a+b)/2*c));
System.out.println("QUADRADO: "+df.format(b*b));
System.out.println("RETANGULO: "+df.format(a*b));
}
}
C++ 代码
#include <cstdio>
int main()
{
double a, b, c;
scanf("%lf%lf%lf", &a, &b, &c);
printf("TRIANGULO: %.3lf\n", a * c / 2);
printf("CIRCULO: %.3lf\n", 3.14159 * c * c);
printf("TRAPEZIO: %.3lf\n", (a + b) * c / 2);
printf("QUADRADO: %.3lf\n", b * b);
printf("RETANGULO: %.3lf\n", a * b);
return 0;
}