程式設計師面試題精選（40）一道SPSS筆試題求解

const double max = numeric_limits::max(); //斜率最大值

class point

float getx() const

float gety() const

//為點重新設定座標

void set(float _x,float _y)

//過載 == 成員操作符

bool operator == (point& p)

//過載流插入操作符

friend istream & operator >> (istream& is, point & p)

};class line //兩點形成一條直線/線段

line(point s,point e):start(s),end(e)

double getk()

};//查詢陣列pp中是否存在點p，是返回陣列序號，否返回-1

int findpoint(point *pp,int size,point &p)

/* p[0].set(0,0);

p[1].set(1,1);

p[2].set(2,0);

p[3].set(1,-1);*/

float left,up,right,down;

//獲取四點x座標最大值和最小值

left = p[0].getx(); //left為四點x座標最小值

right = p[0].getx(); //right為四點x座標最大值

for(i=1;i<4;i++)

//判斷矩形與座標系平**況

point p1(left,up),p2(right,up),p3(right,down),p4(left,down);

if(findpoint(p,4,p1) != -1 && findpoint(p,4,p2) != -1 && findpoint(p,4,p3) != -1 && findpoint(p,4,p4) != -1)

//排序後的四點順時針相連組成矩形邊

line one(*s[0],*s[1]),two(*s[1],*s[2]),three(*s[2],*s[3]),four(*s[3],*s[0]);

cout<<"k1 = "if(one.getk()*two.getk() == -1 || (one.getk() == 0 && two.getk() == max) || (one.getk() == max && two.getk() == 0) )

if(two.getk()*three.getk() == -1 || (two.getk() == 0 && three.getk() == max) || (two.getk() == max && three.getk() == 0) )

if(three.getk()*four.getk() == -1 || (three.getk() == 0 && four.getk() == max) || (three.getk() == max && four.getk() == 0) )

if(four.getk()*one.getk() == -1 || (four.getk() == 0 && one.getk() == max) || (four.getk() == max && one.getk() == 0) )

程式設計師面試題精選（40） 一道SPSS筆試題求解