链接:https://www.nowcoder.com/acm/contest/104/A
来源:牛客网
题目描述
Let be a regualr triangle, and D is a point in the triangle. Given the angle of . Then let AD, CD and BD form a new triangle, what is the size of the three angles?
#include<math.h> #include<stdlib.h> #include<stdio.h> #include<string.h> #include<algorithm> #define pi acos(-1.0) #define eps 1e-12 using namespace std; struct point { double x,y; point(){} point(double x,double y) { this->x=x; this->y=y; } }o1,o2,jiao1,jiao2,p,pp,ppp; double xmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } double Distance(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } //点到直线的距离 double disptoline(point p,point l1,point l2) { return fabs(xmult(p,l1,l2))/Distance(l1,l2); } //求两直线交点 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; } void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2) { point p=c; double t; p.x+=l1.y-l2.y; p.y+=l2.x-l1.x; p=intersection(p,c,l1,l2); t=sqrt(r*r-Distance(p,c)*Distance(p,c))/Distance(l1,l2); p1.x=p.x+(l2.x-l1.x)*t; p1.y=p.y+(l2.y-l1.y)*t; p2.x=p.x-(l2.x-l1.x)*t; p2.y=p.y-(l2.y-l1.y)*t; } void intersection_circle_circle(point c1,double r1,point c2,double r2,point& p1,point& p2) { point u,v; double t; t=(1+(r1*r1-r2*r2)/Distance(c1,c2)/Distance(c1,c2))/2; u.x=c1.x+(c2.x-c1.x)*t; u.y=c1.y+(c2.y-c1.y)*t; v.x=u.x+c1.y-c2.y; v.y=u.y-c1.x+c2.x; intersection_line_circle(c1,r1,u,v,p1,p2); } double c(double A,double B,double C) { return (A*A+B*B-C*C)/2.0/A/B; } int main() { p.x=0.0; p.y=0.0; pp.x=20.0; pp.y=0.0; ppp.x=10.0; ppp.y=10.0*tan(pi/3.0); double A1,A2,A3,r1,r2,A[3],ans[3]; while(~scanf("%lf%lf%lf",&A1,&A2,&A3)) { A1*=pi/180.0; A2*=pi/180.0; A1=(pi-A1)*2.0; A2=(pi-A2)*2.0; o1.x=10.0; o1.y=-10.0/tan(A1/2.0); r1=10.0/sin(A1/2.0); double len=10.0*tan(pi/3.0)+10.0/tan(A2/2.0); o2.x=len*cos(pi/6.0); o2.y=len*sin(pi/6.0); r2=10.0/sin(A2/2.0); intersection_circle_circle(o1,r1,o2,r2,jiao1,jiao2); if(Distance(jiao1,pp)<eps)swap(jiao1,jiao2); A[0]=Distance(jiao1,p); A[1]=Distance(jiao1,pp); A[2]=Distance(jiao1,ppp); sort(A,A+3); //printf("%.9lf %.9lf %.9lf ",A[0],A[1],A[2]); if(A[0]+A[1]>A[2]) { ans[0]=acos(c(A[0],A[1],A[2])); ans[1]=acos(c(A[0],A[2],A[1])); ans[2]=acos(c(A[1],A[2],A[0])); sort(ans,ans+3); for(int i=0;i<3;i++) printf("%.9f ",ans[i]/pi*180.0); printf(" "); } else printf("-1 -1 -1 "); } }
日常学python
while True: try: A=list(map(int,input().split())) A=sorted(A) print(A[0]-60,A[1]-60,A[2]-60) except EOFError: break
str = "Line1-abcdef Line2-abc Line4-abcd"; print str.split( ); print str.split(' ', 1 );