题目链接
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<b<a<105). Separated by an white space.

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

题意描述：
给出一个矩形（给出a，b），需要求出椭圆内接矩形的周长的期望。
看到这里，一个数学小菜鸡掉下了悔恨的眼泪…
高数真滴菜…

在图中，有 $\frac{x^2}{a^2}$+ $\frac{y^2}{b^2}$=1,而y的变化范围是[0,b];
则公式可变换为：
x=a $\sqrt {1-\frac{y^2}{b^2}}$
所以，周长为：
C=4(x+y)=4y+4a $\sqrt {1-\frac{y^2}{b^2}}$
然后就需要求出积分：
$\int_0^b {4y+4a\sqrt {1-\frac{y^2}{b^2}}} \,{\rm d}y$
结果为
=2 $b^2$+ $\pi$ab
然后除以b以后，结果就是
=2b+ $\pi$a

然后加到代码里面求解即可：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double Pi=acos(-1.0);///emmmm，就是pai嘛....
void perimeter()
{
    double a,b;
    scanf("%lf%lf",&a,&b);
    double ans=2*b+Pi*a;///也就是刚刚那个公式
    printf("%.6f\n",ans-0.0000005);///不让进位，进位结果会错的
}
int main()
{
    int t;
    cin >> t;
    while(t--)
        perimeter();
    return 0;
}

GG