题目链接：http://lightoj.com/volume_showproblem.php?problem=1118
给你两个圆的半径和圆心，求交集的面积；
就是简单数学题，但是要注意acos得到的都是小于180度的角，所以这里要注意一下，不要求整个角，求一半的大小；这点让我错的惨不忍睹；
#include <iostream> #include <stdio.h> #include <string.h> #include <string> #include <vector> #include <algorithm> #include <map> #include <queue> #include <stack> #include <math.h> using namespace std; #define met(a, b) memset(a, b, sizeof(a)) #define N 1053 #define INF 0x3f3f3f3f #define PI 4*atan(1) const int MOD = 10000007; typedef long long LL; int main() { int T, t = 1; scanf("%d", &T); while(T--) { double r1, r2, x1, x2, y1, y2; scanf("%lf %lf %lf %lf %lf %lf", &x1, &y1, &r1, &x2, &y2, &r2); double d = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)); if( d >= r1+r2 )///相离或相切 { printf("Case %d: %.6f ", t++, 0.0); continue; } if(d <= fabs(r1-r2))///内含； { printf("Case %d: %.6f ", t++, PI*min(r1, r2)*min(r1, r2)); continue; } double a = acos((r1*r1+r2*r2-d*d)/(2*r1*r2)); double s = sin(a)*r1*r2;///四边形面积 double b = acos((r1*r1+d*d-r2*r2)/(2*r1*d)); double s1 = b * r1 * r1;///r1这边的扇形面积； double c = acos((r2*r2+d*d-r1*r1)/(2*r2*d)); double s2 = c * r2 * r2;///r2这边的扇形面积； double ans = s1 + s2 - s; printf("Case %d: %.7f ", t++, ans); } return 0; } /* Input: 5 0 0 1 10 10 1 0 0 10 0 0 5 -862 823 894 -667 402 663 548 518 145 119 828 620 777 499 712 479 314 967 Output: Case 1: 0.0 Case 2: 78.5398163397 Case 3: 1139058.0639436883 Case 4: 56622.85922574766 Case 5: 1513681.0685423985 */