(CodeForces

(CodeForces - 600D)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.

Examples

Input

0 0 4
6 0 4

Output

7.25298806364175601379

Input

0 0 5
11 0 5

Output

0.00000000000000000000

题目大意：给出两个圆，求出这两个圆重叠的面积。

思路：两个圆可能相离，相交，相含，分类讨论其面积，尤其要注意的是这题需要long double，而且不知道为什么用scanf会wa on test 36……坑的一匹

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>using namespace std;const long double pi=acos(-1.0);int main(){    long double x1,y1,r1,x2,y2,r2;    while(cin>>x1>>y1>>r1>>x2>>y2>>r2)    {        long double d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));        if(r1+r2<=d)         {            printf("0.00000000000000000000\n");            continue;        }        else if(r1>=r2+d||r2>=r1+d)        {            long double r=min(r1,r2);            long double ans=pi*r*r;            printf("%.20f\n",(double)ans);            continue;        }        else        {            long double alpha=acos((r1*r1+d*d-r2*r2)/2/r1/d);            long double beta=acos((r2*r2+d*d-r1*r1)/2/r2/d);            long double ans=(alpha*r1*r1-0.5*r1*r1*sin(alpha*2))+(beta*r2*r2-0.5*r2*r2*sin(beta*2));            printf("%.20f\n",(double)ans);        }    }    return 0;}