hdu 4752

计算几何+数值计算的题目；

要用到辛普森积分，没有学过~~~

参考学习了 acm_Naruto大神 的代码！

代码：

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #define maxn 20005
 5 using namespace std;
 6 int n,a,b,c,l,r,x[maxn],y[maxn];
 7 double sqr(double x)
 8 {
 9     return x*x;
10 }
11 double len(double x)
12 {
13     double t,tt;
14     t=(sqrt((b+2*a*x)*(b+2*a*x)+1)+2*a*x+b);
15     tt=log(t)/(4*a);
16     t=sqrt((b+2*a*x)*(b+2*a*x)+1)*(b+2*a*x)/(4*a);
17     tt+=t;
18     return tt;
19 }
20 
21 int main()
22 {
23     while(scanf("%d%d%d%d%d%d",&n,&a,&b,&c,&l,&r)!=EOF)
24     {
25         for(int i=0; i<n; i++)
26             scanf("%d%d",&x[i],&y[i]);
27         x[n]=x[0],y[n]=y[0];
28         double ans=0;
29         for(int i=0; i<n; i++)if(x[i]!=x[i+1])
30             {
31                 double x0=x[i],y0=y[i];
32                 double x1=x[i+1],y1=y[i+1];
33                 if(x0>x1)swap(x0,x1),swap(y0,y1);
34                 double k=(y1-y0)/(x1-x0);
35                 double d=y1-k*x1;
36                 double u=sqr(b-k)-4*a*(c-d);//b*b-4*a*c;
37                 double length=0;
38                 if(u>=0)
39                 {
40                     double t1=(-(b-k)*1.0+sqrt(u))/(2.0*a);//方程的根;
41                     double t2=(-(b-k)*1.0-sqrt(u))/(2.0*a);//方程的根;
42                     if(t1>t2)swap(t1,t2);
43                     t1=max(x0,max(t1,l*1.0));
44                     t2=min(x1,min(t2,r*1.0));
45                     if(t1<=t2)
46                         length=fabs(len(t2)-len(t1));//辛普森积分；
47                     if(x[i]<x[i+1])ans+=length;
48                     else ans-=length;
49                 }
50             }
51         printf("%.2lf\n",fabs(ans));
52     }
53     return 0;
54 }

View Code

 

 

转载于:https://www.cnblogs.com/yours1103/p/3335955.html