10242 - Fourth Point !!

[_.B._]

Greetings!.
Go to:
http://online-judge.uva.es/board/viewtopic.php?t=5466
It might be of help for you.

P.S.: I used "double" and got AC in 0:00.002, then used "real" and got AC in 0:00.000. The problem says:

All coordinates are between -10000 and +10000.

So I believe that's why "real" is enough.

[mlvahe]

_.B._ I really don't know why there are empty lines in the input.
The only idea I have is there is a blank line after the input.

[_.B._]

Perhaps mlvahe. Might be some mistake if they left the empty line at the end. That's a sure WA for us Pascal users
Still don't know why they don't use simple input, without tricks and whitspaces and empty lines... like I read in one post, they should do a problem of that kind, if they want to, but throw only good input.
Anyway, it was great you noticed it

[Morning]

if i know (x0,y0),(x1,y1),(x2,y2) as three adjacent point of a parallelogram,and (x0,y0) belongs the 1st line,(x1,y1) is the link point of 1st line and 2nd line,(x2,y2) belongs the 2nd line.
can i calculate the 4th point (x3,y3) by

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x3 = x0 + x2 - x1
y3 = y0 + y2 - y1

???????
[cpp]
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
double x0,y0,x1,y1,x2,y2,x3,y3;
while(cin >> x0 >> y0 >> x1 >> y1 >> x1 >> y1 >> x2 >> y2)
{
x3 = x0 + x2 - x1;
y3 = y0 + y2 - y1;
printf("%.3lf %.3lf\n",x3,y3);
}
return 0;
}
[/cpp]

[Eduard]

Formula is right! But nobody tell you that in the input endpoints are given one after another in right form for example your programm can pase this input.

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1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000

But not this.

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1.000 0.000 3.500 3.500 0.000 1.000 3.500 3.500 

Hope it helps.
Eduard

[Morning]

yeah ,u r right.thanks,buddy

[kenneth]

Just a note that for C++ users, use double but NOT float

[apurba]

someone pls explain this problem....no 10242

thanks in advance........

[mf]

What exactly do you want to be explained?

If you don't understand the problem statement, get a dictionary.

[newton]

Why im getting WA?
Please help

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#include <cstdio>
#include <cmath>
#include <algorithm>
#define eps 1e-7

int main(){
	//freopen("in.txt","rt",stdin);
	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(fabs(x2-x3) < eps && fabs(y2-y3) < eps){
			x3 = x4;
			y3 = y4;
		}
		printf("%.3lf %.3lf\n",x1+x3-x2,y1+y3-y2);
	}
	return 0;
}

[Obaida]

You should check which two points match and then go for the 4th point.

[talizmelf]

I checked wich two points match and I still get WA

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#include <iostream>
#include <cstdio>
using namespace std;

int main(){
	double ax1, ay1, ax2, ay2, bx1, by1, bx2, by2, x, y;
	while (!cin.eof()){
		cin >> ax1 >> ay1 >> ax2 >> ay2 >> bx1 >> by1 >> bx2 >> by2;
		if (ax1 == bx1 && ay1 == by1) {x = ax2+bx2-ax1; y = ay2+by2-ay1;}
		else if (ax1 == bx2 && ay1 == by2) {x = ax2+bx1-ax1; y = ay2+by1-ay1;}
		else if (ax2 == bx1 && ay2 == by1) {x = ax1+bx2-ax2; y = ay1+by2-ay2;}
		else {x = ax1+bx1-ax2; y = ay1+by1-ay2;}
		printf("%0.3f %0.3f\n", x, y);
	}
}

can anyone help?

[sh415]

you dont know for sure that (x2,y2) and (x3,y3) are the common point.

its an interesting problem......

here we are give the point of two connected side of a parallelogram.
so; actually we are given 3 points..................
we just have to find the fourth one........

suppose; we are given 1 2 3 2 points(1,2,3 repesents different identity )

here 4th point are same ( 1, 2, 3, 2) with 3 rd point. now we have to decide the 3 rd point from the condition and then calculate the fourth one which is unknown...................

we may have other input like 1 2 1 3 or 1 2 2 3...........................

[live_lie]

thank you keneth and obaida you helped me to get an AC.

[brianfry713]

Input:

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0.000 1.000 0.000 0.000 0.000 1.000 1.000 1.000
0.000 1.000 0.000 0.000 1.000 1.000 0.000 1.000
0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
0.000 0.000 0.000 1.000 1.000 1.000 0.000 1.000

Output should be:

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1.000 0.000
1.000 0.000
1.000 0.000
1.000 0.000