hdoj1071积分求面积
2015-01-20 17:59
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The area
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8022 Accepted Submission(s): 5625
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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).
[align=left]Output[/align]
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
33.33
40.69
Hint
For float may be not accurate enough, please use double instead of float.
代码如下:
#include<stdio.h> #include<math.h> int main() { int t; double x1,y1,x2,y2,x3,y3; double s; scanf("%d",&t); while(t--) { scanf("%lf%lf",&x1,&y1); scanf("%lf%lf",&x2,&y2); scanf("%lf%lf",&x3,&y3); s=(y2-y1)/pow(x2-x1,2)/3*(x3*x3*x3-x2*x2*x2)+(2*(y1-y2)*x1/pow(x2-x1,2)-(y3-y2)/(x3-x2))/2*(x3*x3-x2*x2)+((y2-y1)*x1*x1/pow(x2-x1,2)+y1-y2+x2*(y3-y2)/(x3-x2))*(x3-x2);//分别求二次函数和直线表达式然后相减求积分(要使用顶点坐标设方程能简化过程)过程未知量较多,很难化简,也就懒的去做 printf("%.2f\n",s); } return 0; }
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