1 条题解
-
0
C :
#include"stdio.h" #include"math.h" float bianchang(float x1,float x2,float y1,float y2) { return (sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))); } int main() { float x1,x2,x3,y1,y2,y3; while(1) { scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3); float a,b,c,p; a=bianchang(x1,x2,y1,y2);b=bianchang(x1,x3,y1,y3); c=bianchang(x2,x3,y2,y3); if(a==0&&b==0&&c==0) break; p=(a+b+c)/2; float s; s=sqrt(p*(p-a)*(p-b)*(p-c)); printf("%.1f\n",s); } return 0; }
C++ :
#include<stdio.h> #include<math.h> int main() { float x1,y1,x2,y2,x3,y3,a,b,c,p,s; while(scanf("%f%f%f%f%f%f",&x1,&y1,&x2,&y2,&x3,&y3)) { if(x1==0&&y1==0&&x2==0&&y2==0&&x3==0&&y3==0) break; a=sqrt(1.0*(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)); b=sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2)); c=sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1)); p=(a+b+c)/2.0; s=sqrt(p*(p-a)*(p-b)*(p-c)); printf("%0.1f\n",s); } return 0; }
Java :
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); while(sc.hasNext()) { int x1=sc.nextInt(); int y1=sc.nextInt(); int x2=sc.nextInt(); int y2=sc.nextInt(); int x3=sc.nextInt(); int y3=sc.nextInt(); if(x1==0&&y1==0&&x2==0&&y2==0&&x3==0&y3==0) break; getAreas(x1,y1,x2,y2,x3,y3); } } public static void getAreas(int x1,int y1,int x2,int y2,int x3,int y3) { double ares=Math.abs((double)((x1*y2+x3*y1+x2*y3-x3*y2-x1*y3-x2*y1)/2.0)); System.out.printf("%.1f",ares); System.out.println(); } }
- 1
信息
- ID
- 2132
- 时间
- 3000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者