(CodeForces
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(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;}
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