本文主要是介绍sdut——The area,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
题目描述
Ignatius bought a land last week, but he didn't know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture, can you tell Ignatius the area of the land?
Note: The point P1 in the picture is the vertex of the parabola.
Note: The point P1 in the picture is the vertex of the parabola.
输入
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).
输出
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.
示例输入
2 5.000000 5.000000 0.000000 0.000000 10.000000 0.000000 10.000000 10.000000 1.000000 1.000000 14.000000 8.222222
示例输出
33.33 40.69
#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{int t;double x1,y1,x2,y2,x3,y3,a,b,c,K,B;scanf("%d",&t);while(t--){cin>>x1>>y1>>x2>>y2>>x3>>y3;a=(y2-y1)/((x2-x1)*(x2-x1));b=(y2-y3-(a*(x2*x2-x3*x3)))/(x2-x3);c=y2-a*x2*x2-b*x2;K=(y2-y3)/(x2-x3);B=y2-K*x2;// double sum1=(a/3)*(pow(x3,3.0)-pow(x2,3.0))-b/2*(pow(x3,2.0)-pow(x2,2.0))-c*(x3-x2);// double sum2=K/2*(x3*x3-x2*x2)-B*(x3-x2);double ans1=((a/3)*x3*x3*x3+(b/2)*x3*x3+c*x3)-((a/3)*x2*x2*x2+(b/2)*x2*x2+c*x2);double ans2=((K/2)*x3*x3+B*x3)-((K/2)*x2*x2+B*x2);//cout<<fixed<<setprecision(2)<<(sum1-sum2)<<endl;cout<<fixed<<setprecision(2)<<(ans1-ans2)<<endl;}return 0;
}
#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{int t;double x1,y1,x2,y2,x3,y3,a,b,c,K,B;scanf("%d",&t);while(t--){cin>>x1>>y1>>x2>>y2>>x3>>y3;a=(y2-y1)/((x2-x1)*(x2-x1));b=(y2-y3-(a*(x2*x2-x3*x3)))/(x2-x3);c=y2-a*x2*x2-b*x2;K=(y2-y3)/(x2-x3);B=y2-K*x2;// double sum1=(a/3)*(pow(x3,3.0)-pow(x2,3.0))-b/2*(pow(x3,2.0)-pow(x2,2.0))-c*(x3-x2);// double sum2=K/2*(x3*x3-x2*x2)-B*(x3-x2);double ans1=((a/3)*x3*x3*x3+(b/2)*x3*x3+c*x3)-((a/3)*x2*x2*x2+(b/2)*x2*x2+c*x2);double ans2=((K/2)*x3*x3+B*x3)-((K/2)*x2*x2+B*x2);//cout<<fixed<<setprecision(2)<<(sum1-sum2)<<endl;cout<<fixed<<setprecision(2)<<(ans1-ans2)<<endl;}return 0;
}
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