下载数据的问题-正方形的顶点

Question

我想编写一个程序，在其中给出正方形的两个顶点，然后程序找到另外两个顶点。

例如：

输入：

2 3 4 5

输出

（（3,6）（0,5）

但是当我从程序中读取数据时，我遇到了问题，一切正常

int main(void)
{

    rationalNumber *z,e,f,g,h;

    point *x, a, b, c, d;

    e.l = 2; e.m = 1; f.l=3; f.m=1; g.l = 4; g.m = 1; h.l=5; h.m=1;

    a.x = e;
    a.y = f;
    b.x = g;
    b.y = h;
    output_point(D_2(z,a, b));


return 0;
}

但是当从用户读取数据时，我收到此消息

“线程1：EXC_BAD_ACCESS（代码= 1，地址= 0x0）”

在这个地方：

if (scanf("%d/%u",&(a->n), &(a->d)) == 1) a->d=1;

所有代码：

#include<stdio.h>
#include<stdlib.h>

typedef struct rn{
    int n;   /**numerator**/
    unsigned d;    /**denomirator**/
} rationalNumber;

int gcd(int,int);

void output(rationalNumber);
void input(rationalNumber *);
void Minus(rationalNumber*, rationalNumber, rationalNumber);
void Sum( rationalNumber *, rationalNumber, rationalNumber);
void Multiplication(rationalNumber *, rationalNumber, rationalNumber);
void Reciprocal( rationalNumber *);

typedef struct dot{
    rationalNumber x;
    rationalNumber y;
} point;

void load_point(point *);
void output_point(point);
point C_1(rationalNumber *, point, point);
point D_1(rationalNumber *, point, point);
point C_2(rationalNumber *, point, point);
point D_2(rationalNumber *, point, point);


int main(void)
{

    rationalNumber *z;
    point *a,b,c;

    load_point(a);
    b = *a;
    load_point(a);
    c = *a;

    output_point(D_2(z,b, c));


return 0;
}

int gcd(int a, int b)
{
  if(b!=0)
    return gcd(b,a%b);

       return a;
}


void output(rationalNumber a)
{
    if(a.d == 1)  printf("%d",a.n);
    else printf("%d/%u",a.n, a.d);
}

void input(rationalNumber *a)
{
    int nwd;
    if (scanf("%d/%u",&(a->n), &(a->d)) == 1) a->d=1;
    else
    {
        nwd = abs(gcd(a->n, a->d));
        a->n = a->n/nwd;
        a->d = a->d/nwd;
    }


}

void Minus(rationalNumber *a, rationalNumber b, rationalNumber c)
{
   int nwd;
      if(b.d == c.d)
      {
          a->n = b.n - c.n;
          a->d = b.d;
          nwd = abs(gcd(a->n, a->d));
          a->n = a->n/nwd;
          a->d = a->d/nwd;

      }else
      {
          a->n = b.n*c.d - c.n*b.d;
          a->d = b.d * c.d;
          nwd = abs(gcd(a->n, a->d));
          a->n = a->n/nwd;
          a->d = a->d/nwd;

      }
}

rationalNumber minus( rationalNumber a)
{
    return a;
}

void Sum( rationalNumber *a, rationalNumber b, rationalNumber c)
{
    int nwd;
    if(b.d == c.d)
    {
        a->n = b.n + c.n;
        a->d = b.d;
    }else
    {
        a->n = b.n*c.d + c.n*b.d;
        a->d = b.d * c.d;
        nwd = abs(gcd(a->n, a->d));
        a->n = a->n/nwd;
        a->d = a->d/nwd;

    }

}

void Multiplication( rationalNumber *a, rationalNumber b, rationalNumber c)
{
    int nwd;

    a->n = b.n*c.n;
    a->d = b.d *c.d;

    nwd = abs(gcd(a->n, a->d));
    a->n = a->n/nwd;
    a->d = a->d/nwd;

}


void Reciprocal( rationalNumber *a)
{
    int buffor;

    if(a->n<0)
    {
        buffor = abs(a->n);
        a->n = -(a->d);
        a->d = buffor;
    }
    else if(a->n>0)
    {
        buffor = a->n;
        a->n = a->d;
        a->d = buffor;
    }


}

void load_point(point *a)
{
    input(&a->x);
    input(&a->y);

}

void output_point(point a)
{
    printf("(");
    output(a.x);
    printf(",");
    output(a.y);
    printf(")\n");
}

point C_1(rationalNumber *c_1, point a, point b)
{

    point c;

    Sum(c_1,b.x,a.y);

    Minus(c_1,*c_1, b.y);
    c.x = *c_1;

    Sum(c_1,b.y,b.x);

    Minus(c_1,*c_1, a.x);
    c.y = *c_1;

    return c;
}

point D_1(rationalNumber *d_1, point a, point b)
{
    point d;

    Sum(d_1,a.x,a.y);

    Minus(d_1,*d_1, b.y);
    d.x = *d_1;

    Sum(d_1,a.y,b.x);

    Minus(d_1,*d_1, a.x);
    d.y = *d_1;

    return d;
}

point C_2(rationalNumber *c_1,point a, point b)
{
    point c;

    Minus(c_1,b.x,a.y);

    Sum(c_1,*c_1, b.y);
    c.x = *c_1;

    Minus(c_1,b.y,b.x);

    Sum(c_1,*c_1, a.x);
    c.y = *c_1;

    return c;
}

point D_2(rationalNumber *d_1,point a, point b)
{
    point d;

    Minus(d_1,a.x,a.y);

    Sum(d_1,*d_1, b.y);
    d.x = *d_1;

    Minus(d_1,a.y,b.x);

    Sum(d_1,*d_1, a.x);
    d.y = *d_1;

    return d;
}

我不知道该如何解决

Answer 1

[如果正方形的一侧为(a, b)，则可以立即将其两个法线之一作为(b, -a)（或(-b, a)，您的问题是模棱两可的），最后一点可以通过将两个向量相加而得到，因此(a + b, b - a)（或(a - b, b + a)，取决于您首先选择的矢量）

要转移到平面上任何一对点的唯一方法是先减去一个矢量以移动图形原点...。解决问题后，您可以将整个对象移回图形的原点。点到了。

因此，假设给定两个向量P = (a, b)和Q = (c, d)，第一个任务是从P = (a, b)中减去Q = (c, d)以使其从原点开始，然后得到P'(a - a, b - b) = P'(0, 0)； Q' = (c - a, d - b)，现在生成法线向量（交换坐标并将符号更改为其中一个），比​​方说R' = (-d + b, c - a)，然后将两个向量相加得到S' = (c - a - d + b, c - a + d - b)，即对角线相反。这是您得到的两点。让我们将整个内容移回原始点：

P' = (0, 0) ==> P = (0 + a, 0 + b) = (a, b) = P
Q' = (c - a, d - b) ==> (c - a + a, d - b + b) = (c, d) = Q
R' = (b - d, c - a) ==> (b - d + a, c - a + b) = R
S' = (c - a - d + b, c - a + d - b) ==> (c - a - d + b + a, c - a + d - b + b) = (c + b - d, c - a + d) = S

因此，如果得到两个点P = (a, b)和Q = (c, d)，则必须返回R = (a + b - c, -a + b + c)和S = (b + c - d, -a + c + d)。

double a, b, c, d;

scan("%g%g%g%g", &a, &b, &c, &d);
printf("Given (%g, %g) and (%g, %g),\n"
       "we got the other two vertices "
       "as (%g, %g) and (%g, %g)\n",
       a, b, c, d,
       a + b - c, b - a + c, 
       b + c - d, c - a + d);