HDU 1071

http://acm.hdu.edu.cn/showproblem.php?pid=1071

推定积分公式。。。。

设直线方程:y=kx+t…………………………………………………………(1)
  抛物线方程:y=ax^2+bx+c……………………………………………………(2)
已知抛物线顶点p1(x1,y1),两线交点p2(x2,y2)和p3(x3,y3)
斜率k=(y3-y2)/(x3-x2)……………………………………………………(3)
把p3点代入(1)式结合(3)式可得:t=y3-(k*x3)
又因为p1是抛物线的顶点,可得关系:x1=-b/2a即b=-2a*x1………………(4)
把p1点代入(2)式结合(4)式可得:a*x1*x1-2a*x1*x1+c=y1化简得c=y1+a*x1*x1……(5)
把p2点代入(2)式结合(4)式和(5)式可得:a=(y2-y1)/((x1-x2)*(x1-x2))
于是通过3点求出了k,t,a,b,c即两个方程式已求出
题目时求面积s
通过积分可知:s=f(x2->x3)(积分符号)(ax^2+bx+c-(kx+t))
               =f(x2->x3)(积分符号)(ax^2+(b-k)x+c-t)
               =[a/3*x^3+(b-k)/2*x^2+(c-t)x](x2->x3)
               =a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3-(a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2)
化简得:
面积公式:s=-(y2-y1)/((x2-x1)*(x2-x1))*((x3-x2)*(x3-x2)*(x3-x2))/6;

#include<stdio.h>
int main()
{
    double a,b,c,t,k,x1,x2,x3,y1,y2,y3,s1,s2;
    int te;
    scanf("%d",&te);
    while(te--)
    {
       scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
       k=(y3-y2)/(x3-x2);
       t=y3-k*x3;
       a=(y2-y1)/((x1-x2)*(x1-x2));
       b=-x1*2*a;
       c=y1-a*x1*x1-b*x1;
       s1=1.0/3*a*x2*x2*x2+1.0/2*(b-k)*x2*x2+x2*(c-t);
       s2=1.0/3*a*x3*x3*x3+1.0/2*(b-k)*x3*x3+x3*(c-t);
       printf("%.2lf
",s2-s1);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/huzhenbo113/p/3254644.html