题意:
给出两个圆的圆心坐标和半径，求这两个圆的公切线切点的坐标及对应线段长度。若两圆重合，有无数条公切线则输出-1.
输出是按照一定顺序输出的。
分析：
首先情况比较多，要一一判断，不要漏掉。
如果高中的那点老底还在的话，代码还是很好理解的。
1 //#define LOCAL 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 using namespace std; 8 9 const double PI = acos(-1.0); 10 const double EPS = 1e-8; 11 struct Point 12 { 13 double x, y; 14 Point(double x=0, double y=0):x(x), y(y) {} 15 16 }; 17 typedef Point Vector; 18 Vector operator + (Vector A, Vector B) 19 { 20 return Vector(A.x+B.x, A.y+B.y); 21 } 22 Vector operator - (Vector A, Vector B) 23 { 24 return Vector(A.x-B.x, A.y-B.y); 25 } 26 Vector operator * (Vector A, double p) 27 { 28 return Vector(A.x*p, A.y*p); 29 } 30 Vector operator / (Vector A, double p) 31 { 32 return Vector(A.x/p, A.y/p); 33 } 34 double dcmp(double x) 35 { 36 if(fabs(x) < EPS) return 0; 37 else return x < 0 ? -1 : 1; 38 } 39 bool operator < (const Vector& a, const Vector& b) 40 { 41 return dcmp(a.x-b.x) < 0 || dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) < 0; 42 } 43 bool operator == (const Vector& a, const Vector& b) 44 { 45 return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; 46 } 47 double Dot(Vector a, Vector b) 48 { 49 return a.x*b.x + a.y*b.y; 50 } 51 double Cross(Vector a, Vector b) 52 { 53 return a.x*b.y - a.y*b.x; 54 } 55 double Length(Vector a) 56 { 57 return sqrt(Dot(a, a)); 58 } 59 struct Circle 60 { 61 double x, y, r; 62 Circle(double x, double y, double r):x(x), y(y), r(r) {} 63 Point point(double a) 64 { 65 return Point(x + r*cos(a), y + r*sin(a)); 66 } 67 }; 68 int getTangents(Circle A, Circle B, Point* a, Point* b) 69 { 70 int cnt = 0; 71 if(A.r < B.r) { swap(A, B); swap(a, b); } 72 double d2 = (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y); 73 double rdiff = A.r - B.r; 74 double rsum = A.r + B.r; 75 if(d2 < rdiff*rdiff) return 0; //内含 76 77 double base = atan2(B.y-A.y, B.x-A.x); 78 if(dcmp(d2) == 0 && dcmp(A.r - B.r) == 0) return -1; //重合 79 if(dcmp(d2 - rdiff*rdiff) == 0) //内切 80 { 81 a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++; 82 return 1; 83 } 84 85 //有外公切线 86 double ang = acos((A.r - B.r) / sqrt(d2)); 87 a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++; 88 a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++; 89 if(dcmp(rsum*rsum - d2) == 0) 90 {//外切 91 a[cnt] = b[cnt] = A.point(base); cnt++; 92 } 93 else if(dcmp(d2 - rsum*rsum) > 0) 94 { 95 ang = acos((A.r + B.r) / sqrt(d2)); 96 a[cnt] = A.point(base + ang); b[cnt] = B.point(PI + base + ang); cnt++; 97 a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); cnt++; 98 } 99 return cnt; 100 } 101 102 int main(void) 103 { 104 #ifdef LOCAL 105 freopen("10674in.txt", "r", stdin); 106 #endif 107 108 int x1, y1, r1, x2, y2, r2; 109 while(scanf("%d%d%d%d%d%d", &x1, &y1, &r1, &x2, &y2, &r2) == 6 && r1 && r2) 110 { 111 Point a[4], b[4]; 112 Circle C1(x1, y1, r1), C2(x2, y2, r2); 113 int n = getTangents(C1, C2, a, b); 114 printf("%d\n", n); 115 int p[4] = {0, 1, 2, 3}; 116 for(int i = 0; i < n; ++i) 117 for(int j = i+1; j < n; ++j) 118 if(a[p[j]] < a[p[i]] || (a[p[j]] == a[p[i]] && b[p[j]] < b[p[i]])) swap(p[i], p[j]); 119 for(int i = 0; i < n; ++i) 120 printf("%.5lf %.5lf %.5lf %.5lf %.5lf\n", a[p[i]].x, a[p[i]].y, b[p[i]].x, b[p[i]].y, Length(a[p[i]] - b[p[i]])); 121 } 122 123 return 0; 124 }