oval-and-rectangle

oval-and-rectangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 445    Accepted Submission(s): 198
 

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

Source

2018 Multi-University Training Contest 6

解析：可以参考这篇博客https://blog.csdn.net/codeswarrior/article/details/81515593

比赛时候只知道题意，根本没想到用积分做，而且这道题需要舍去第六位小数后面所有的位数，但是printf这个输出时候是按照四舍五入的舍去的，因此我又了解了下各种取整的库函数，这片博客中https://blog.csdn.net/xiamentingtao/article/details/52593648写的很详细，floor、round、ceil、fix这四种的区别，，，，这道题可以先乘以1e6,用floor函数向下取整，再除以1e6即可。

代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define pi acos(-1)
int main()
{
    int t;
    double a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf",&a,&b);
        double sum=(2.0*b+a*pi)*1e6;
        sum=floor(sum);
        printf("%.6lf\n",sum/1e6);
    }
    return 0;
}