gaussian elimination calculator

", 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

saw discuss ... OK or do both sides, the first time normal run, the second time the greedy selection of small, and the demand behind the point can not affect the front point.Here is still to do the blog comments:1. Why ca=b instead of ac=b??? (I don't understand it at first)Because each robot corresponds to a row vector, and we want the row vectors in a to correspond to the other line vectors in B, it is obvious that the direct moment multiplication is not possible, but if we transpose a a A, a

Because the state of the previous row is only relevant to the current row So we can consider enumerating the 1 And then start from line 2nd to line 6th, depending on the state of the previous row. Time Complexity is O ((1 The code is as follows: #include The method of Gaussian elimination is to list 30 equations and to find out 30 unknowns by these equations. And here the method of constructing the equ

It feels pretty simple. Because the Gaussian elimination in the I, J Subscript hit the reverse, kneeling a few hair ... We can build a equations that resemble a[i]^x1^x2^...xn = B[i] Then we can solve the equation. The i,j indicates that the first I switch will affect the J lamp It means that the first J lamp is affected by XI, do not reverse the The code is as follows: #include

http://www.lydsy.com/JudgeOnline/problem.php?id=1013As long as the equations are listed, we can set Gauss to solve them.Obviously the distance is equal, so it doesn't matter if the square is open.b represents the center point, which can be listedSigma ((X[i][j]-b[j]) ^2) =sigma ((X[i+1][j]-b[j]) ^2)To simplifySigma (2*b[j]* (x[i+1][j]-x[i][j)) =sigma (x[i+1][j]^2-x[i][j]^2)Then we get n equations, and the problem is guaranteed to have a solution, just set Gauss on the line.The first time to lear

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

Bestcoder Round #90At least three knowledge points exploded at this time ....A. ZZ questionPress Test instructions copy init function and then count OKB.The game problem does not understand the half-day of the SG ..... The result of this question ....CData structure problem I wrote for half an hour and found that the color was changed .... Oh, my God, deep-fried.D.I don't know what to do with the topic ... The official puzzle is to kill the little circle.Bestcoder Round #90//div All-in-one roll-

The main idea: the tree to pull the light gameGaussian extinction solution or equation set, for all free element violence 2^n enumeration state, substituting computationIt's a bit of an elegant approach ...#include Bzoj 2466 Zhongshan 2009 Tree Gaussian elimination + violence

; jvar+1; J + +) A[i][j]^=A[k][j]; } } } for(intI=k; i) if(A[i][col])return-1; if(kvar)return var-K; for(intI=var-1; i>=0; i--) {X[i]=a[i][var]; for(intj=i+1; jvar; J + +) X[i]^= (a[i][j]X[j]); } return 0;}intn,m;voidinit () {memset (A,0,sizeof(a)); memset (x,0,sizeof(x)); Equ=n*m; var=n*m; for(intI=0; i) for(intj=0; j) { intt=i*m+J; A[t][t]=1; if(i>0) a[(I-1) *m+j][t]=1; if(i1) a[(i+1) *m+j][t]=1; if(j>0) a[i*m+j-1][t]=1; if(j1) a[i*m+j+1][t]

smallest x of the equation, then the matrix A in the above equation can be represented by a larger matrix n*n, and then it is OK to convert the relationship (specifically in the code). So now is the n unknown, n equations, in (1), you can add l to both sides, then it becomes: x (All) *a (all) + x (*a) 1,3 (*a) + x (1,3) 2,1 (*a) + ... + x (3 , 3) *a (3,3) = L (2) x * a = L (like this) Then we're enumerating all the states: the first thing we want to enumerate is the number of

What is Gaussian elimination method? See Wikipedia Definition MATLAB implementation: Function x = Gauss (A, B)N = length (B );For I = 1: n-1For J = (I + 1): nIf a (I, I )~ = 0Lam = a (J, I)/A (I, I );A (J, (I + 1): n) = a (J, (I + 1): N)-lam * a (I, (I + 1 ): n );B (j) = B (j)-lam * B (I );EndEndEndFor I = N:-1:1B (I) = (B (I)-sum (A (I, (I + 1): N ). * B (I + 1): n ))). /A (I, I );EndX = B; Python im

The main idea: give a non-right graph, find a path from 1 to N, so that the value of the path on the difference or maximum, the path can be repeated.InputThe first line contains two integers n and M, indicating the number of points in the graph and the number of edges. The next M-line describes M-bar, with three integers per line si,ti, Di, which indicates that there is a non-forward edge between Si and Ti with a weighted value of Di. There may be heavy edges or self-loops in the diagram.OutputC

The main topic: Beg Σ[i|n]i^dOnlookers: http://www.cnblogs.com/jianglangcaijin/p/4033399.htmlSure enough, I'm still too konjac--In addition, the 0 items of σ[1#include Bzoj 36,011 person's number theory of the Mo-de-si inverse + Gaussian elimination