oval-and-rectangle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 237 Accepted Submission(s): 93
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.
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.
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.
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
Source
Recommend
题意：给你椭圆的短轴长和长轴长，求y=c和椭圆相交的点的矩形周长的期望
分析：首先根据题目给出的椭圆短轴长和长轴长可以得出椭圆的方程：x^2/a^2 + y^2/b^2
然后把y=c带进椭圆的方程可以求出对应的横坐标，矩形的周长即4*c+4*a*sqrt(1-c*c/b*b)
接着对椭圆的周长公式进行积分，注意我们求的是周长的期望所以积分的结果还要除以b（y可以取的最大值）
积分过程：
上面图片最后两行公式错了，应该是2b*b+π*a*b，最后除以b以后的期望是2*b+π*a
AC代码：
#pragma comment(linker, "/STACK:102400000,102400000")
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e6+10;
const double PI = acos(-1.0);
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int a,b,t;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
cin>>t;
while(t--) {
cin>>a>>b;
double ans = 2.0*b+PI*a;
///double ans = 1.0*a*a+2.0*a/b*PI;
ans = ans-5*1e-7; //严格控制在小数点后六位
printf("%.6lf\n",ans);
///printf("%.6lf\n",ans);
}
return 0;
}