elimination tournament generator

A[k][i] 0 - for(k=i+1; k) +a[k][j]-=a[k][i]/a[i][i]*A[i][j]; A } at for(i=n-1; i>=0; i--) {//Back to Generation - for(j=i+1; j) -a[i][n]-=a[j][n]*A[i][j]; -A[i][n]/=A[i][i]; - } - } in - intMain () { toscanf"%d",n); + for(intI=0; i for(intj=0; j) -scanf"%LF",a[i][j]); the for(intI=0; i//Tectonic equation set * for(intj=0; j2* (a[i+1][j]-a[i][j]); $ for(intj=0; j1][j]*a[i+1][j]-a[i][j]*A[i][j];Panax Notoginseng } - Gause (); the

Notoginseng if(p!=i) - WAPs (p,i); the if(a[i].d[col]==0) + { Acol++; thei--; + Continue; - } $ for(intj=i+1; jj) $ { - Doublenow=a[i].d[col]/A[j].d[col]; - for(intk=col;k1;++k) the { -a[j].d[k]*=now,a[j].d[k]-=A[i].d[k];Wuyi } the } -col++; Wu } - } About voidGet_ans () $ { - for(intI=n;i>0;--i) - { - for(intj=n;j>i;--j) A { +a[i].d[n+1]-=a[i].d[j]*Ans[j];

0.0Sample Output0.500 1.500HINTData size:对于40%的数据，1Tips:给出两个定义：1、 球心：到球面上任意一点距离都相等的点。2、 距离：设两个n为空间上的点A, B的坐标为(a1, a2, …, an), (b1, b2, …, bn)，则AB的距离定义为：dist = sqrt( (a1-b1)^2 + (a2-b2)^2 + … + (an-bn)^2 )ExercisesGaussian elimination of the naked problem.According to the hint can be used as a point of entry, because the n+1 points, so you can use a point as a datum point, and other n points to form n equations.Code:#include #include #include #include

reduced to 0 by F[i][i], and the other elements of the peers are proportionally subtracted from the corresponding values of the first row - //that is, the elementary line transformation of matrices + } A } atFord (I,n,1){ - DoubleT= (f[i][n+1]/=f[i][i]); -Foru (J,1, I-1) f[j][n+1]-= (T*f[j][i]);//eliminate Yuan - } - } - in intMain () { -scanf"%d",n); toForu (I,1, N) scanf ("%LF",r[i]); +Foru (I,1, N) { -Foru (J,1, N) { thescanf"%LF",x); *f[i][j]=2* (xr[j]);

rows greater than line I from the main element of this column. It is important to note that the main element should not be 0 and not very small, otherwise it will seriously affect the accuracy (that is, if it is small, it cannot reflect the denominator). So we are going to find the largest of column I, and then exchange it with the current line at each elimination, so that we can get the maximum main element.The problem of determining the non-solutio

", stdin); -Freopen ("Bzoj_1013.out","W", stdout); the #endif * intN; $scanf"%d",n);Panax Notoginseng for(intI=1; i1; i++) - for(intj=1; j) thescanf"%LF",a[i][j]); + A for(intI=1; i){ the for(intj=1; j){ +a[i][n+1]-=a[i][j]*a[i][j]-a[i+1][j]*a[i+1][j]; -a[i][j]=2* (a[i+1][j]-a[i][j]); $ } $ } - - Solve (n); the - for(intI=1; i)Wuyiprintf"%.3LF", a[i][n+1]); theprintf"\ n"); - return 0; Wu}Linear al

var; J + +) { if(SGN (a[i][j])! =0Free_x[j]) free_x_num+ +, Free_index =J; } if(Free_x_num >1)Continue; Temp= a[i][var]; for(j =0; J var; J + +) { if(SGN (a[i][j])! =0 J! =free_index) Temp-= a[i][j] *X[j]; } X[free_index]= temp/A[i][free_index]; Free_x[free_index]=0; } return var-K; } for(i =var-1; I >=0; i--) {Temp= a[i][var]; for(j = i +1; J var; J + +) { if(SGN (a[i][j])! =0) Temp-= a[i][j] *X[j]; } X[i]= temp/A[i][i]; } re

DescriptionThere 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 OutputinputThe first line is an integer n (1OutputWith only one row, the n-dimensio

limit:1 Sec Memory limit:162 MBsubmit:4166 solved:2191[Submit] [Status] [Discuss] DescriptionThere is a spherical space generator capable of producing a hard sphere in n-dimensional space. Now that you're trapped in this n-dimensional sphere, you only know the ball.The coordinates of the n+1 points on the surface, you need to determine the spherical coordinates of the n-dimensional sphere as quickly as you can to destroy the sphere space gen

Topic:http://www.lydsy.com/JudgeOnline/problem.php?id=1013ExercisesConsider the two-dimensional we can understand a truth:Two points to the left can represent an equation, and then subtract from the 22 equation to get a one-time equationSo we can do it with Gaussian elimination.#include #include#include#include#defineN 13using namespacestd;intn,m;DoubleC[n][n],f[n][n],ans[n];inlinevoidGauss () { for(intI=1; i) { intL=i; for(intj=l+1; j)

data, 1For 100% of data, 1Tip: Give two definitions:1, the ball sphere: to the spherical surface any point distance is equal points.2, Distance: Set two n for Space point A, B coordinates (a1, a2, ..., an), (B1, B2, ..., BN), then AB distance is defined as: dist = sqrt ((A1-B1) ^2 + (A2-B2) ^2 + ... + (an-bn) ^2)Ps:Gaussian elimination!Set the center of the circle (x, y), and then use the radius equal! Using the distance formula we can get n equation

Subtract the upper and lower two equations and then Gaussian elimination 1013: [JSOI2008] spherical space generator sphere time limit: 1 Sec Memory Limit: 162 MB Submit: 2901 Solved: 1517 [Submit] [Status] [Discuss] DescriptionThere is a spherical space generator capable of producing a hard sphere in n-dimensional space. Now tha

Topic: Given n-dimensional space under the n+1 point, the N-point of the spherical sphereI've been trying for a long time. Simulated annealing 0.0 still not AC 0.0Today, digging the dung wall to learn the Gaussian elimination yuan ...We set the sphere to X (x1,x2,..., xn)Suppose there are two points a (a1,a2,..., an) and B (B1,b2,..., bn)Then we can get two equations.(X1-A1) ^2+ (X2-A2) ^2+...+ (Xn-an) ^2=r^2(X1-B1) ^2+ (X2-B2) ^2+...+ (xn-bn) ^2=r^2T

- { - Doublegg=ga[j][i-1]/ga[i-1][i-1]; in for(intk=i;k1; k++)//several unknown items of the equation -ga[j][k]=gg*ga[i-1][k]-Ga[j][k]; to } + Solve (); - } the * intMain () $ {Panax Notoginsengscanf"%d",n); - for(intI=1; i1; i++) the for(intj=1; j) +scanf"%LF",f[i][j]); A for(intI=1; i) the for(intj=1; j) +Ga[i][j]= (f[i+1][J]-F[I][J]) *2, ga[i][n+1]+=f[i+1][j]*f[i+1][j]-f[i][j]*F[i][j]; - Guass (); $ for(intI=1; i) $pr

Transmission DoorTitle DescriptionThere is a spherical space generator capable of producing a hard sphere in the nnn dimension space. Now that you are trapped in this nnn sphere, you only know the coordinates of the n+1n+1n+1 points on the sphere, and you need to determine the spherical coordinates of the nnn sphere as quickly as you can to destroy the sphere space generator.Input/output formatInput format:The first line is an integer nnn (1Output for

According to the simple column equation, we can list n+1 of n-element 2-th equation.By subtracting the adjacent two equations, we can get N-n-ary 1-th equation, and the Gaussian elimination element can be done.1 varB:Array[0.. -,0.. -] ofextended;2Temp,ans:Array[0.. -] ofextended;3 I,j,k,n:longint;4 cnt:extended;5 begin6 READLN (n);7 fori:=1 ton+1 Do8 forj:=1 toN Do9 read (b[i,j]);Ten fori:=1 toN Do One forj:=1 toN Do

Seek the god of ~>_Based on the relationship of equal radii, the n+1 two-time equations are established.Then each and the previous subtract two entries, get n a linear equation.#include Gaussian elimination bzoj1013 [JSOI2008] spherical space generator sphere