codeforce--600D

题意：求相交圆的面积。借鉴大神代码，精度超高。

#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#include <ctime>
#include <assert.h>

#define y1 aasdfasdfasdf

#define eps 1e-16
#define M_PI 3.141592653589793
const int N = 200005;
using namespace std;

long double x1,y1,x2,y2,r1,r2;

long double gd(long double x1,long double y1,long double x2,long double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

long double solve_cos(long double a,long double b,long double c)
{
    return acos((a*a+b*b-c*c)/(2*a*b));
}

long double cut(long double ang,long double r)
{
    long double s1,s2;
    s1=ang*r*r/2;
    s2=sin(ang)*r*r/2;
    return s1-s2;
}

long double solve()
{
    if (r1<r2)
    {
        swap(x1,x2);
        swap(y1,y2);
        swap(r1,r2);
    }
    long double cd=gd(x1,y1,x2,y2);
    if (cd+r2<=r1+eps)
        return r2*r2*M_PI;
    if (cd>=r1+r2-eps)
        return 0;
    long double ang1=solve_cos(cd,r1,r2);
    long double ang2=solve_cos(cd,r2,r1);
    return cut(ang1*2,r1)+cut(ang2*2,r2);
}

int main()
{
    cin>>x1>>y1>>r1;
    cin>>x2>>y2>>r2;
    cout.precision(12);
    cout<<fixed<<solve()<<endl;
    return 0;
}