The areaTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8190Accepted Submission(s): 5748
Problem Description
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.
Input
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).
Output
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.
Sample Input
25.000000 5.0000000.000000 0.00000010.000000 0.00000010.000000 10.0000001.000000 1.00000014.000000 8.222222
Sample Output
33.3340.69
Hint
For float may be not accurate enough, please use double instead of float.
主要就是利用高数中的积分知识~
代码：#include <stdio.h>#include <math.h>#define deta 0.01struct Point{double x , y ;};int main(){int t ;scanf("%d",&t);while(t–){Point p1,p2,p3;scanf("%lf%lf",&p1.x,&p1.y) ;scanf("%lf%lf",&p2.x,&p2.y) ;scanf("%lf%lf",&p3.x,&p3.y) ;double k = (p3.y-p2.y)/(p3.x-p2.x) ;if(p2.x>p3.x){Point t = p2;p2 = p3 ;p3 = t ;}double A,C;C = p1.y ;A = (p3.y-C)/((p3.x-p1.x)*(p3.x-p1.x)) ;double ans = 0.0 ;for(double i = p2.x+deta ; i <= p3.x ; i+=deta){double y = fabs(A*(i-p1.x)*(i-p1.x)+C-k*(i-p2.x)-p2.y) ;ans += y*deta ;}printf("%.2lf\n",ans) ;}return 0 ;}与君共勉
，世上再美的风景，都不及回家的那段路。
