zoukankan      html  css  js  c++  java
  • 求平行四边形的第四个点的坐标

    Given are the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. Find the (x,y) coordinates of the fourth point.
    Input
    Each line of input contains eight floating point numbers: the (x,y) coordinates of one of the endpoints of the first side followed by the (x,y) coordinates of the other endpoint of the first side, followed by the (x,y) coordinates of one of the endpoints of the second side followed by the (x,y) coordinates of the other endpoint of the second side. All coordinates are in meters, to the nearest mm. All coordinates are between -10000 and +10000.
    Output
    For each line of input, print the (x,y) coordinates of the fourth point of the parallelogram in meters, to the nearest mm, separated by a single space.
    Sample Input
    0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
    1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
    1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
    
    Sample Output
    1.000 0.000
    -2.500 -2.500
    0.151 -0.398
    
    题意 :给你平行四边形的相邻的两条边,让你求第四个点的坐标
    思路分析:一个比较基础的计算几何题,只要根据向量的矢量和相等即可求解
        相等的点的位置有 4 种情况,分别是 1 3、 1 4、 2 3、 2 4分别求即可
    代码示例 :

    struct point
    {
        double x, y;
        point(double _x=0, double _y=0):x(_x), y(_y){}
         
        // 点-点=向量
        point operator-(const point &v){
            return point(x-v.x, y-v.y);
        }
     
    };
     
    int dcmp(double x){
        if (fabs(x)<eps) return 0;
        else return x<0?-1:1;
    }
    bool operator == (const point &a, const point &b){
        return (dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0);
    }
     
    typedef point Vector; // Vector表示向量
    
    point p[10];
    
    int main() {
        //freopen("in.txt", "r", stdin);
        //freopen("out.txt", "w", stdout);
        
        while(~scanf("%lf%lf", &p[1].x, &p[1].y)){
            for(int i = 2; i <= 4; i++) {
                scanf("%lf%lf", &p[i].x, &p[i].y);
            }
            
            if (p[2] == p[3]) {
                Vector v = p[4] - p[3];
                printf("%.3f %.3f
    ", v.x+p[1].x, v.y+p[1].y);
            }
            if (p[2] == p[4]){
                Vector v = p[3] - p[4];
                printf("%.3f %.3f
    ", v.x+p[1].x, v.y+p[1].y);
            }
            if (p[1] == p[3]){
                Vector v = p[4]-p[3];
                printf("%.3f %.3f
    ", v.x+p[2].x, v.y+p[2].y);
            }
            if (p[1] == p[4]){
                Vector v = p[3] - p[4];
                printf("%.3f %.3f
    ", v.x+p[2].x, v.y+p[2].y);
            }
        }
        return 0;
    }
    
    东北日出西边雨 道是无情却有情
  • 相关阅读:
    hdu4578线段树维护平方和,立方和(加,乘,赋值)或者珂朵莉树
    珂朵莉树(ODT老司机树)
    Codeforces Round #524 (Div. 2)D
    HDU1402 FFT高精度乘法模板题
    中国剩余定理poj1006
    POJ
    Install and Config MySQL 8 on Ubuntu
    Protobuf Examples
    Learning Thrift
    Flask Quickstart
  • 原文地址:https://www.cnblogs.com/ccut-ry/p/8995055.html
Copyright © 2011-2022 走看看