HDOJ 1071 The area (纯数学题)
//题目地址
//http://acm.hdu.edu.cn/showproblem.php?pid=1071
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 int main() 6 { 7 int T; 8 double x1,x2,x3,y1,y2,y3,a,b,c,k,t,x,s1,s2; 9 cin>>T; 10 while(T--) 11 { 12 cin>>x1>>y1>>x2>>y2>>x3>>y3; 13 a=(y2-y1)/((x1-x2)*(x1-x2)); 14 b=-2*a*x1; 15 c=y1+a*x1*x1; 16 k=(y3-y2)/(x3-x2); 17 t=y3-(k*x3); 18 s1=a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3; 19 s2=a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2; 20 printf("%.2lf\n",s1-s2); 21 } 22 return 0; 23 } 24 /* 25 设直线方程:y=kx+t…………………………………………………………(1) 26 抛物线方程:y=ax^2+bx+c……………………………………………………(2) 27 已知抛物线顶点p1(x1,y1),两线交点p2(x2,y2)和p3(x3,y3) 28 斜率k=(y3-y2)/(x3-x2)……………………………………………………(3) 29 把p3点代入(1)式结合(3)式可得:t=y3-(k*x3) 30 又因为p1是抛物线的顶点,可得关系:x1=-b/2a即b=-2a*x1………………(4) 31 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) 32 把p2点代入(2)式结合(4)式和(5)式可得:a=(y2-y1)/((x1-x2)*(x1-x2)) 33 于是通过3点求出了k,t,a,b,c即两个方程式已求出 34 题目时求面积s 35 通过积分可知:s=f(x2->x3)(积分符号)(ax^2+bx+c-(kx+t)) 36 =f(x2->x3)(积分符号)(ax^2+(b-k)x+c-t) 37 =[a/3*x^3+(b-k)/2*x^2+(c-t)x](x2->x3) 38 =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) 39 化简得: 40 面积公式:s=-(y2-y1)/((x2-x1)*(x2-x1))*((x3-x2)*(x3-x2)*(x3-x2))/6; 41 */