"JSOI2008" Spherical space generator

P2182-"JSOI2008" Spherical space generator description

There is a spherical space generator capable of producing a hard sphere in n-dimensional space. Now that you are trapped in this n-dimensional sphere, you only know the coordinates of the n+1 points on the sphere, and you need to determine the spherical coordinates of the n-dimensional sphere as quickly as you can to destroy the sphere space generator.

Input

The first line is an integer, N.
The next n+1 line, each row has n real numbers, representing the n-dimensional coordinates of a point on the sphere. Each real number is accurate to 6 digits after the decimal point, and its absolute value is no more than 20000.

Output

With only one row, the n-dimensional coordinates (n real numbers) of the globe are given in turn, and two real numbers are separated by a space.
Each real number is exactly 3 digits after the decimal point. Data is guaranteed to be solvable. Your answer must be the same as the standard output in order to score.

Sample Input

2
0.0 0.0
-1.0 1.0
1.0 0.0

Sample Output

0.500 1.500

Hint

Data size:

For 40% of data, 1<=n<=3

For 100% of data, 1<=n<=10

Tip: Give two definitions:

1, the ball sphere: to the spherical surface any point distance is equal points.

2, Distance: Set two n for the space point A, B coordinates (a1, a2, ..., an), (B1, B2, ..., bn), the distance of AB is defined as:
dist = sqrt ((A1-B1) ^2 + (A2-B2) ^2 + ... + (an-bn) ^2)

Analysis x=3 The situation:

set the sphere to x, Y, radius is R . 

You can list the equations:

(a1-x) ^2+ (b1-y) ^2+ (c1-z) ^2=r^2;

(a2-x) ^2+ (b2-y) ^2+ (c2-z) ^2=r^2;

(a3-x) ^2+ (b3-y) ^2+ (c3-z) ^2=r^2;

(a4-x) ^2+ (b4-y) ^2+ (c4-z) ^2=r^2;

by simplifying each equation and subtracting equation 1 from the 2,3,4 equation, we can get 3 equations that are similar to this one:

2 (A2-A1) +2 (B2-B1) +2 (C2-C1) =a2^2-a1^2+b2^2-b1^2+c2^2-c1^2 . 

In this way, the two-time equation is reduced to one-time equation, then the Gaussian elimination method is used to solve the equation.

Gaussian elimination element method:

Put the coefficients in a matrix, and finally the matrix into this form:

1	0	0	X
0	1	0	Y
0	0	1	Z

finally x, y is the answer. 

Enumerates each column to eliminate the element.

① Select a main element:intt=i;while (!a[t][i]) t++;

② The main element to the current line. 

③ the coefficients of the principal elements into one. 

④ then the coefficients of the other equations on this column of the main element are converted to 0. (Plus and minus the elimination element method)

1#include <Set>2#include <map>3#include <queue>4#include <stack>5#include <ctime>6#include <cmath>7#include <string>8#include <vector>9#include <cstdio>Ten#include <cstdlib> One#include <cstring> A#include <iostream> -#include <algorithm> - using namespacestd; the Doublea[ the][ the],b[ the][ the]; - intN; - voidGauss () { -    for(intI=1; i<=n;i++){ +     intt=i; -      while(!b[t][i]) t++; +      for(intj=1; j<=n+1; j + +) Swap (b[i][j],b[t][j]); A     Doublek=B[i][i]; at      for(intj=i;j<=n+1; j + +) b[i][j]/=K; -      for(intj=1; j<=n;j++) -       if(J!=i &&B[j][i]) { -k=B[j][i]; -      for(intp=i;p<=n+1;p + +) -b[j][p]-=k*B[i][p]; in       } -   } to } + intMain () - { theFreopen ("!. Inch","R", stdin); *Freopen ("!. out","W", stdout); $scanf"%d",&n);Panax Notoginseng    for(intI=1; i<=n+1; i++) -      for(intj=1; j<=n;j++) thescanf"%LF",&a[i][j]); +    for(intI=2; i<=n+1; i++)//column Equation A      for(intj=1; j<=n;j++) theb[i-1][j]= (a[i][j]-a[1][J]) *2, b[i-1][n+1]+= (a[i][j]*a[i][j]-a[1][j]*a[1][j]); + Gauss (); -    for(intI=1; i<=n;i++) $printf"%.3LF", b[i][n+1]); $   return 0; -}

"JSOI2008" Spherical space generator