#include<iostream.h>
#include<conio.h>
#include<fstream.h>
#include<iomanip.h>
#include<math.h>
//int wdf();   
///
//这个程序还有很大问题的
//
//
///
int Loa1=585,Loa2=1050,Lba=500,Lae=1800,Lbe=2300,Lbc=450,Lde=450;   //这么多全局变量合适吗？？
int Lcd1=2600,Lcd2=2300,Lcd3=2300,Lcd4=2600;
int Loaj=-10,Loax=10,Lcdzheng=10,Lcdfu=-10;
int Loa1i,Loa2i,i=1,B;
int k1,k2,k3;
int Lcd11,Lcd22,Lcd33,Lcd44;
int Xa,Xb=-400,Xc=-400,Ya,Yb=800,Yc=1250,Xe,Ye,Xd,Yd;
int jiaodu_fai=atan2((-Xb),Yb),jiaodu_ED,jiaodu_beita,jiaodu_aerfa;      //角度定义
//int cosB=(Xb*Xb+Yb*Yb+Lba*Lba+-Loa1i*Loa1i)/(2*Lba*sqrt(Xb*Xb+Yb*Yb)); 
//
int a,b;   //计算角度B的中间变量，没什么用
int wtf(int m,int n)
{
Loa1i=m;
Lcd11=n;
a=sqrt(  4*Lba*Lba*(Xb*Xb+Yb*Yb)-(Xb*Xb+Yb*Yb+Lba*Lba-Loa1i*Loa1i)*(Xb*Xb+Yb*Yb+Lba*Lba-Loa1i*Loa1i)  );   /
b=Xb*Xb+Yb*Yb+Lba*Lba-Loa1i*Loa1i;                              
jiaodu_beita=atan2(a,b);
Xa=Xb+Lba*sin(B+jiaodu_fai);
Ya=Yb+Lba*cos(B+jiaodu_fai);
jiaodu_aerfa=atan2(Ya,Xa);
//
Xe=Xb+(Lba+Lae)*sin(B);
Ye=Yb+(Lba+Lae)*cos(B);
k1=2*(Ye-Yc)*Lde;
k2=2*(Xe-Xc)*Lde;
k3=(Xe-Xc)*(Xe-Xc)+(Ye-Yc)*(Ye-Yc)+Lde*Lde-Lcd11*Lcd11;                     //
jiaodu_ED=2*atan2((k1+sqrt(k1*k1+k2*k2-k3*k3)),(k2-k3));
Xd=Xe+Lde*cos(jiaodu_ED);
Yd=Ye+Lde*sin(jiaodu_ED);
cout<<"第 "<<i<<" 个:  "<<"Xd: "<<Xd<<"  Yd: "<<Yd<<endl;
cout<<endl;
return 0;
};
void main()//难死你妈的了
{
Loa1i=Loa1+i*Loaj;   // -10
Loa2i=Loa2+i*Loax;   // +10
Lcd11=Lcd1+i*Lcdfu;
Lcd22=Lcd2+i*Lcdzheng;
Lcd33=Lcd3+i*Lcdfu;
Lcd44=Lcd4+i*Lcdzheng;
Lcd1=2600;
//装载机工作机构从1到2到3到4一共完成3个过程，其余不予考虑//
///从1位置到2位置的过程Lcd..减小/
for(i;i<=30;i++)  
{
Lcd11=Lcd1+i*Lcdfu;
wtf(Lcd11,Loa1);
}
/
//从2位置到3位置Lcd..的长度不变 Loa..增加/
/*	for(1;1<30;i++)
{	
Loa2i=Loa2i+i*Loax;
wtf();
}*/
//
//从位置3到位置4Loa..不变，Lcd..增加//
for(1;1<30;i++)
{	
Lcd22=Lcd2+i*Lcdzheng;
}
//
}