1 条题解
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0
C :
#include<stdio.h> #include<math.h> int main() { float a,b,c,d; scanf("%f%f%f",&a,&b,&c); d=(a+b+c)/2.; d=sqrt(d*(d-a)*(d-b)*(d-c)); printf("%.3f",d); }C++ :
#include<stdio.h> #include<math.h> int main() { double a,b,c,s,p; while(scanf("%lf %lf %lf",&a,&b,&c)!=EOF) { p=(a+b+c)/2.0000; p=p*(p-a)*(p-b)*(p-c); s=sqrt(p); printf("%.3lf\n",s); } }Java :
import java.util.Scanner; public class Main { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner input = new Scanner(System.in); double a = input.nextDouble(); double b = input.nextDouble(); double c = input.nextDouble(); double s = (a + b + c)/2.0; double area = Math.sqrt(s*(s-a)*(s-b)*(s-c)); System.out.printf("%.3f",area); input.close(); } }Python :
a = map(lambda x:float(x), raw_input().split()) x = sum(a)/2 print "%.3f"%((x*(x-a[0])*(x-a[1])*(x-a[2]))**0.5)
- 1
信息
- ID
- 2022
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- (无)
- 递交数
- 0
- 已通过
- 0
- 上传者