题目的链接如下：http://poj.org/problem?id=2624
4th Point Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5057 Accepted: 1760
Description 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
题目的大意如下：给你三个点ABC （我假设输入中给出的相同的那个点是点A)让你求平行四边形的第四个点D（注意这里的三个点不一定就是第二个点和第三个点相同),然后用向量做的话：AB + AC = AD 所以D点的坐标就是B+C-A的坐标，具体的代码如下：
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #define eps 1e-8 struct TPoint { double x,y; TPoint operator-(TPoint&a){ TPoint p1; p1.x=x-a.x; p1.y=y-a.y; return p1; } }; using namespace std; int main() { TPoint ans; TPoint a[10]; while(~scanf("%lf%lf", &a[0].x, &a[0].y)) { ans.x = a[0].x, ans.y = a[0].y; int flag = 1, t = 0; for(int i = 1; i < 4; i++) { scanf("%lf%lf", &a[i].x, &a[i].y); ans.x += a[i].x; ans.y += a[i].y; for(int j = 0 ; flag && j < i; j++){ if(a[i].x == a[j].x && a[i].y == a[j].y){ flag = 0; t = i; } } } ans.x -= 3 * a[t].x; ans.y -= 3 * a[t].y; printf("%.3f %.3f\n", ans.x, ans.y); } return 0; }