sgu233：The Greatest Angle(计算几何)

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题目大意：
给出一个圆O和严格在圆内的两点A,B，在圆上求一点C使得∠ACB最大。

分析：
首先我们得知道A,B,C所构成的外接圆与圆O相切。如果O到AB的垂线垂足位于AB中点使得有两个切点的话我们考虑离AB近的那一个。(不知道请走隔壁数学组)
知道这个之后就是乱搞了…(本蒟蒻参考的刘汝佳的做法)
http://wenku.baidu.com/link?url=DYfmAczQnMSS2aXUkQdLQ7bkNl7yLq05nfLElMDu249JWIqHhytV3kQg_T36tw3dCOBJeYYIS3Yj-9Y-QfSYQuTGQT9ic9g0FPeYcAtxQ8O

AC code：

#include <cstdio>#include <cmath>#include <algorithm>typedef double DB;#define ONLINE_JUDGEusing namespace std;const DB eps = 1e-8;const DB pi = acos(-1.0);struct pot{    DB x, y;    pot(DB x = 0, DB y = 0):x(x),y(y){}    void read() {scanf("%lf%lf", &x, &y);}    void print() {printf("%.6lf %.6lf", x, y);}    DB length() {return sqrt(x*x+y*y);}    friend pot operator + (const pot &a, const pot &b) {return pot(a.x+b.x, a.y+b.y);}     friend pot operator - (const pot &a, const pot &b) {return pot(a.x-b.x, a.y-b.y);}    friend pot operator * (const pot &a, DB k) {return pot(a.x*k, a.y*k);}    friend pot operator / (const pot &a, DB k) {return pot(a.x/k, a.y/k);}    pot unit() {return pot(x, y)/length();}}O, A, B, C, vCP, vAC, oP, D;DB R, r;DB AC, CO, cosPCO, OCA;DB P, Q;DB a, b, c, deita;DB PC;DB p, q;DB sqr(DB x) {return x*x;}DB dis2(const pot &a, const pot &b) {return sqr(a.x-b.x)+sqr(a.y-b.y);}DB dis(const pot &a, const pot &b) {return sqrt(dis2(a, b));}bool is_zero(DB x){    if(x >= -eps && x <= eps) return true;    return false;}DB calc_angle(const pot &a, const pot &b, const pot &c){    DB ac2 = dis2(a, c), bc2 = dis2(b, c), ab2 = dis2(a, b);    if(is_zero(bc2) || is_zero(ab2)) return 0;    return acos((ac2-bc2-ab2)/2/sqrt(bc2)/sqrt(ab2));}bool is_correct(const pot &P){    if(is_zero(dis(P, O)+r-R)) return true;    return false;   }int main(){    #ifndef ONLINE_JUDGE    freopen("sgu233.in", "r", stdin);    freopen("sgu233.out", "w", stdout);    #endif    int test;    scanf("%d", &test);    while(test--)    {        O.read(), scanf("%lf", &R);        A.read(), B.read(), C = (A+B)/2;        AC = dis(A, C), CO = dis(C, O);        OCA = calc_angle(A, C, O);        cosPCO = cos(OCA+pi/2);        P = sqr(R)-dis2(O, C)+dis2(A, C);        Q = 2*CO*cosPCO;        a = sqr(Q)-4*sqr(R), b = 2*P*Q, c = sqr(P)-4*dis2(A, C)*sqr(R);        if(is_zero(a)) PC = -c/b;        else        {            deita = sqr(b)-4*a*c;            if(is_zero(deita)) deita = 0;            p = -b/2/a, q = sqrt(deita)/2/a;            PC = p+q;            if(sqr(p+q) > sqr(p-q)) PC = p-q;        }        r = sqrt(sqr(PC)+dis2(A, C));        vAC = C-A, vAC = vAC/vAC.length();        vCP = pot(-vAC.y, vAC.x), PC = fabs(PC);        if(is_correct(C+vCP*PC)) oP = C+vCP*PC;        else oP = C-vCP*PC;        D = (oP-O).unit()*R+O;        D.print();puts("");    }    #ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);    #endif    return 0;}

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