题意：给你镜子的位置（用两点确定的一条直线表示），光源，光的反射点，求光在镜子的折射点
计算几何的模板，注意斜率！！！
模板1：
#include<cstdio>#include<stdlib.h>#include<string.h>#include<string>#include<map>#include<cmath>#include<iostream>#include <queue>#include <stack>#include<algorithm>#include<set>using namespace std;#define INF 1e8#define eps 1e-4#define ll __int64#define maxn 500010#define mol 1000000007struct point { double x,y;};point symmetric_point(point p1, point l1, point l2)// 求p1的对称点{ point ret; if (l1.x > l2.x - eps && l1.x < l2.x + eps)//斜率不存在 { ret.x = (2 * l1.x - p1.x); ret.y = p1.y; } else { double k = (l1.y - l2.y ) / (l1.x - l2.x); if(k + eps > 0 && k - eps < 0)//斜率为零 { ret.x = p1.x; ret.y = l1.y - (p1.y - l1.y); } else { ret.x = (2*k*k*l1.x + 2*k*p1.y - 2*k*l1.y - k*k*p1.x + p1.x) / (1 + k*k); ret.y = p1.y - (ret.x - p1.x ) / k; } } return ret;}point intersection(point u1,point u2,point v1,point v2)//求两直线的交点{ point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret;}int main(){ int t; scanf("%d",&t); while(t--) { point m1,m2,l1,l2; scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&m1.x,&m1.y,&m2.x,&m2.y,&l1.x,&l1.y,&l2.x,&l2.y); point tmp=symmetric_point(l1,m1,m2); point ans=intersection(tmp,l2,m1,m2); printf("%.3lf %.3lf\n",ans.x,ans.y); } return 0;}模板2（利用点到直线上的最近点避免斜率）：
#include<cstdio>#include<cmath>using namespace std;const double eps=1e-8;struct point{ double x,y;};point intersection(point u1,point u2,point v1,point v2){ point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret;}point ptoline(point p,point l1,point l2){ point t=p; t.x+=l1.y-l2.y,t.y+=l2.x-l1.x; return intersection(p,t,l1,l2);}point m1,m2,l1,l2;int main(){ int n; scanf("%d",&n); while(n--){ scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&m1.x,&m1.y,&m2.x,&m2.y,&l1.x,&l1.y,&l2.x,&l2.y); point dot = ptoline(l1,m1,m2); point now; now.x = 2*dot.x - l1.x;//源点与对称点的中点落在直线上 now.y = 2*dot.y - l1.y; point ans = intersection(now,l2,m1,m2); printf("%.3lf %.3lf\n",ans.x,ans.y); } return 0;}