HDU6362 oval-and-rectangle(2018HDU多校联赛第六场，数学,积分)

Problem Description

Patrick Star find an oval.
The half of longer axes is on the x-axis with length a.
The half of shorter axes is on the y-axis with length b.
Patrick Star plan to choose a real number c randomly from [0,b], after that, Patrick Star will get a rectangle :
\1. The four vertexes of it are on the outline of the oval.
\2. The two sides of it parallel to coordinate axis.
\3. One of its side is y=c.
Patrick Star want to know the expectations of the rectangle’s perimeter.

Input

The first line contain a integer T (no morn than 10), the following is T test case, for each test case :
Each line contains contains two integer a, b (0

Output

For each test case output one line denotes the expectations of the rectangle’s perimeter .
You should keep exactly 6 decimal digits and ignore the remain decimal digits.
It is guaranted that the 7-th decimal digit of answer wont be 0 or 9.

Sample Input

1
2 1

Sample Output

8.283185

思路

留下了数学渣的泪水…

题目给了一个椭圆,然后让你求内接矩形的周长的期望，(a,b$a,b$已知)

做法是先把椭圆的周长加起来再除以变化范围。

比如这个图，我们知道焦点在x$x$轴的椭圆的公式为x2a2+y2b2=1$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,而y$y$的变化范围是[0,b]$[0,b]$,那么由公式变形可以知道:

x2=a2∗(1−y2b2)$x^2=a^2\ast (1-\frac{y^2}{b^2})$

x=a1−y2b2−−−−−√$x=a\sqrt{1-\frac{y^2}{b^2}}$

然后就可以知道椭圆上的点的坐标为(a1−y2b2−−−−−√,y)$(a\sqrt{1-\frac{y^2}{b^2}},y)$

那么周长为:

C=4(x+y)=4y+4a1−y2b2−−−−−√$C=4(x+y)=4y+4a\sqrt{1-\frac{y^2}{b^2}}$

然后就需要求出积分:

∫by=0(4∗y+4a∗1−y2b2−−−−−√)dy${\int }_{y=0}^b(4\ast y+4a\ast \sqrt{1-\frac{y^2}{b^2}})dy$

换元，设y=bsin(α)$y=bsin(\alpha )$, dy=bcos(α)dα$dy=bcos(\alpha )d\alpha$.

原式 =∫1/2πα=0(4b∗sin(α)+4a∗cos(α))bcos(α)dα$={\int }_{\alpha =0}^{1/2\pi }(4b\ast sin(\alpha )+4a\ast cos(\alpha ))bcos(\alpha )d\alpha$

=∫1/2πα=0(2b2∗sin(2α)+2ab∗(cos(2α)+1))dα$={\int }_{\alpha =0}^{1/2\pi }(2b^2\ast sin(2\alpha )+2ab\ast (cos(2\alpha )+1))d\alpha$

=[−b2cos(2α)+ab(sin(2α)+2α)]1/2π0$=[-b^{2}cos(2\alpha )+ab(sin(2\alpha )+2\alpha ){]}_0^{1/2\pi }$

=[b2+πab]−[−b2]$=[b^2+\pi ab]-[-b^2]$

=2b2+πab$=2b^2+\pi ab$

因为变化范围是[0,b]$[0,b]$,所以最后算期望除以 b$b$ 。

=2b+πa$=2b+\pi a$

代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
const double pi=acos(-1.0);
const int N=1e5+10;
void solve()
{
    double a,b;
    scanf("%lf%lf",&a,&b);
    double ans=2*b+pi*a;
    printf("%.6f\n",ans-0.0000005);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}