1 条题解
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0
C :
#include<stdio.h> #include<math.h> int main(){ double a,b,c,s,L; scanf("%lf %lf %lf",&a,&b,&c); s=(a+b+c)/2; L=sqrt(s*(s-a)*(s-b)*(s-c)); printf("%.2lf",L); return 0; }
C++ :
#include <stdio.h> #include <math.h> int main() { float a, b, c, s, area; scanf("%f %f %f", &a, &b, &c); s = (a + b + c) * 0.5; area = sqrt(s * (s - a) * (s - b) * (s - c)); printf("%.2f\n", area); return 0; }
Pascal :
var a,b,c,d,s:double; begin readln(a,b,c); d:=(a+b+c)/2; s:=sqrt(d*(d-a)*(d-b)*(d-c)); writeln(s:0:2); end.
Java :
import java.util.*; public class Main { public static void main(String args[]) { Scanner cin=new Scanner(System.in); double a,b,c; double p; double hl; a=cin.nextDouble(); b=cin.nextDouble(); c=cin.nextDouble(); p=(a+b+c)/2; hl=Math.sqrt(p*(p-a)*(p-b)*(p-c)); System.out.printf("%.2f",hl); } }
Python :
from math import sqrt a,b,c = [float(x) for x in raw_input().split()] p = (a+b+c)/2 print "%.2f" %sqrt(p*(p-a)*(p-b)*(p-c))
- 1
信息
- ID
- 1247
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者