3 条题解
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812209水宇辰 (12209水宇辰) LV 9 @ 2021-11-28 11:20:59
#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main() { float X1,Y1,X2,Y2,X3,Y3,p; float a,b,c; cin >>X1>>Y1>>X2>>Y2>>X3>>Y3; a=sqrt(pow((X1-X2),2)+pow((Y1-Y2),2)); b=sqrt(pow((X1-X3),2)+pow((Y1-Y3),2)); c=sqrt(pow((X2-X3),2)+pow((Y2-Y3),2)); p=(a+b+c)/2; cout<<fixed<<setprecision(2)<<sqrt(p*(p-a)*(p-b)*(p-c)); return 0; }
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-12024-10-27 10:43:02@
#include<bits/stdc++.h>
using namespace std;
int main()
{
double x1,y1,x2,y2,x3,y3;
cin>>x1>>y1>>x2>>y2>>x3>>y3;
cout<<setprecision(2)<<fixed<<fabs(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))/2;
return 0;
} -
-62021-11-24 19:30:53@
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
float x1,x2,x3,y1,y2,y3,a,b,c,p,s;
cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
a = sqrt(pow((x1-x2),2)+pow((y1-y2),2));
b = sqrt(pow((x2-x3),2)+pow((y2-y3),2));
c = sqrt(pow((x3-x1),2)+pow((y3-y1),2));
p = (a+b+c)/2;
s = sqrt(p*(p-a)*(p-b)*(p-c));
cout << fixed << setprecision(2) << s << endl;
return 0;
}//就差一个就对了,帮我看看题解发下
- 1
信息
- ID
- 2326
- 难度
- 4
- 分类
- (无)
- 标签
- 递交数
- 179
- 已通过
- 82
- 通过率
- 46%
- 被复制
- 2
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