[UVA][Math] 10242 - Fourth Point !!
Fourth Point!!
Input: standard input
Output: standard output
Time Limit: 2 seconds
Memory Limit: 32 MB
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. Input is terminated by end of file.
OutputFor 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 Input0.000 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
判斷平行四邊形的第四個點在哪,
首先會知道兩個邊一定會有一個共同的點, 這樣就會有答案了
#include <stdio.h>
int main() {
double x1, y1, x2, y2, x3, y3, x4, y4;
while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf",
&x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4) == 8) {
if(x1 == x3 && y1 == y3)
printf("%.3lf %.3lf\n", (x2+x4)-x1, (y2+y4)-y1);
else if(x1 == x4 && y1 == y4)
printf("%.3lf %.3lf\n", (x2+x3)-x1, (y2+y3)-y1);
else if(x2 == x3 && y2 == y3)
printf("%.3lf %.3lf\n", (x1+x4)-x2, (y1+y4)-y2);
else
printf("%.3lf %.3lf\n", (x1+x3)-x2, (y1+y3)-y2);
}
return 0;
}