Codeforces Educational Round 2 D题

D. Area of Two Circles' Intersection

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two circles. Find the area of their intersection.

Input

The first line contains three integers x1, y1, r1 ( - 109 ≤ x1, y1 ≤ 109, 1 ≤ r1 ≤ 109) — the position of the center and the radius of the first circle.

The second line contains three integers x2, y2, r2 ( - 109 ≤ x2, y2 ≤ 109, 1 ≤ r2 ≤ 109) — the position of the center and the radius of the second circle.

Output

Print the area of the intersection of the circles. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Sample test(s)

input

0 0 4
6 0 4

output

7.25298806364175601379

input

0 0 5
11 0 5

output

0.00000000000000000000

题意：很简单，给两个圆，求其相交部分的面积。

解法：推公式，具体公式在这里暂时就不给出了（留个坑），需要注意精度问题，赛中用之前的模板一发过了，CF上大神就是多，我这个模板过了2014年北京区域赛的题，结果居然被hack掉了，说明之前的模板精度有问题，下面是新的模板，精度应该是非常可靠了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define esp 1e-8

using namespace std;

struct Circle{
    long double x,y;
    long double r;
};

long double calArea(Circle c1, Circle c2)
{
    long double d;
    long double s,s1,s2,s3,s4,angle1,angle2,temp;

    d=sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
    //printf("%.10f\n",d);
    if(d>=(c1.r+c2.r))//相离
        return 0;
    if((c1.r-c2.r)>=d)//内含
        return acos(-1.0)*c2.r*c2.r;
    if((c2.r-c1.r)>=d)//内含
        return acos(-1.0)*c1.r*c1.r;
    angle1=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
    angle2=acos((c2.r*c2.r+d*d-c1.r*c1.r)/(2*c2.r*d));
    s1=angle1*c1.r*c1.r;
    s2=angle2*c2.r*c2.r;
    s3=c1.r*c1.r*sin(angle1)*cos(angle1);
    s4=c2.r*c2.r*sin(angle2)*cos(angle2);
    s=s1+s2-s3-s4;
    return s;
}

Circle a,b;

int main(){
    while(cin>>a.x>>a.y>>a.r){
        cin>>b.x>>b.y>>b.r;
        cout.precision(12);
        cout<<calArea(a,b)<<endl;
    }
    return 0;
}