zoukankan      html  css  js  c++  java
  • The area (hdu1071)积分求面积

    The area

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7066 Accepted Submission(s): 4959


    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
    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
     
    Sample Output
    33.33
    40.69
    Hint
    For float may be not accurate enough, please use double instead of float.
     
     
     
    一道高数题。
     
    回顾一下微积分的知道,曲面的微积分。两个方程相减,对x或者y积分,就是面积。
    然后具体的公式详见代码把,
     
     
    #include<stdio.h>
    //#include<math.h>与y1方法重载了、
    double x1,y1,x2,y2,x3,y3;
    double area;
    double l,h,a,k,b;
    
    //曲线的方程可用顶点式  y = a(x-h)^2+l, (h,l) 为顶点坐标)
    //直线方程  y = kx+b;
    
    //图形一定,只是点的位置不定。所以对x积分
    double f(double x)
    {
        return (a*x*x*x/3)-(a*h+k/2)*x*x+(a*h*h+l-b)*x;//积分,化简
    }
    int main()
    {
    
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
            h=x1;
            l=y1;
            a=(y2-y1)/((x2-h)*(x2-h));
            k=(y3-y2)/(x3-x2);
            b=y2-k*x2;
            printf("%.2lf
    ",f(x3)-f(x2));
        }
        return 0;
    }
  • 相关阅读:
    D365FO Debug找不到w3cp进程
    D365FO 10.0PU32 开发环境 Data Management导出失败
    一张图看懂项目管理
    用户体验为什么重要?如何提升产品的用户体验?(写给产品小白)
    敏捷考证?你应该知道的敏捷体系认证(最全名单)
    漫画:禅道程序员的一天
    敏捷开发管理--任务分解经验之谈
    漫画:优秀程序员的必备特质有哪些?
    漫画:女生/男生告白攻略
    漫画:程序员脱单秘籍
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/3636870.html
Copyright © 2011-2022 走看看