Circular Area
Time Limit: 1000MS Memory Limit: 65536K
Description
Your task is to write a program, which, given two circles, calculates the area of their intersection with the accuracy of three digits after decimal point.
Input
In the single line of input file there are space-separated real numbers x1 y1 r1 x2 y2 r2. They represent center coordinates and radii of two circles.
Output
The output file must contain single real number - the area.
Sample Input
20.0 30.0 15.0 40.0 30.0 30.0
Sample Output
608.366
Source
Northeastern Europe 2000, Far-Eastern Subregion
这道题直接上模板就可以了~
模板：
struct Circle{ double x, y, r; }; //圆的圆心坐标，半径 double dis(Circle a, Circle b){ return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } //两圆圆心的距离 double solve(Circle a, Circle b){ double d = dis(a, b); if (d >= a.r + b.r) return 0; if (d <= fabs(a.r - b.r)){ double r = a.r < b.r ? a.r : b.r; return pi * r * r; } double ang1 = acos((a.r * a.r + d * d - b.r * b.r) / 2.00 / a.r / d); double ang2 = acos((b.r * b.r + d * d - a.r * a.r) / 2.00 / b.r / d); double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1); return ret; } //返回值即为两圆公共部分的面积
POJ2546：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define pi 3.1415926535897932384626 6 7 struct Circle{ 8 double x, y, r; 9 } r[31]; 10 11 double dis(Circle a, Circle b){ 12 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); 13 } 14 15 double solve(Circle a, Circle b){ 16 double d = dis(a, b); 17 if (d >= a.r + b.r) return 0; 18 if (d <= fabs(a.r - b.r)){ 19 double r = a.r < b.r ? a.r : b.r; 20 return pi * r * r; 21 } 22 23 double ang1 = acos((a.r * a.r + d * d - b.r * b.r) / 2.00 / a.r / d); 24 double ang2 = acos((b.r * b.r + d * d - a.r * a.r) / 2.00 / b.r / d); 25 double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1); 26 return ret; 27 } 28 29 int main(){ 30 31 while (~scanf("%lf%lf%lf%lf%lf%lf", &r[0].x, &r[0].y, &r[0].r, &r[1].x, &r[1].y, &r[1].r)) 32 printf("%.3f ", solve(r[0], r[1])); 33 34 return 0; 35 36 }