gaussian elimination calculator

Test instructionsA rectangular area is divided into m*n units numbered (1, 1) to (M, N), upper left (1, 1), and lower right (M, N). Given P (k) i,j, wherein 1≤i≤m,1≤j≤n,1≤k≤4, indicates the probability of (I, J) to (I+1, J), (I, j+1), (I-1, J), (I, j-1). A knight in (1, 1), followed by a given probability, each step is irrelevant before asking to reach (M, N) the desired step number.ResolutionIt's easy to thinkThen move the item to write the determinantFigure cut the arrogant guyThe probability

outputcase #1:3Case #2:3 Test Instructions: It is to take at least one number inside these numbers so that the product of these numbers is a complete square number; Idea: A number is the total square number, that it decomposition of prime numbers when the power of each prime factor is even, x[i]={0,1} indicates the number of selected or not selected, then for the J-Factor, ∑x[i]*a[i][j]=0 (MOD2); A[i][j]A recurses that represents the number of J primes in the first, and then obtains a matrix, t

Question: hdoj 4975 a simple Gaussian elimination problem. This question is the same as that of hdoj 4888, but the data has been enhanced. This question is indeed not good, especially the data. It is highly controversial, but it also shows that optimization is sometimes very useful. If you do not understand, you can refer to this explanation: Click This question only adds a bit of optimization, that is, if

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4975 A simple Gaussian elimination problem.Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)Total submission (s): 669 accepted submission (s): 222Problem descriptiondragon is studying math. one day, he drew a table with several rows and columns, randomly wrote numbers on each elements of the table. then he counted the sum of eac

. OutputFor each puzzle, the output consists of a line with the string: "puzzle # m", where M is the index of the puzzle in the input file. following that line, is a puzzle-like display (in the same format as the input ). in this case, 1's indicate buttons that must be pressed to solve the puzzle, while 0 indicate buttons, which are not pressed. there shoshould be exactly one space between each 0 or 1 in the output puzzle-like display. Sample Input 20 1 1 0 1 01 0 0 1 1 10 0 1 0 0 11 0 0 1 0 10

POJ 3185 The Water Bowls (Gaussian elimination method, enumeration free variable), pojbowls Question: The Water Bowls Time Limit:1000 MS Memory Limit:65536 K Total Submissions:5013 Accepted:1960 DescriptionThe cows have a line of 20 water bowls from which they drink. the bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down

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 format:With only one row, the nnn coordinates of

Topic PortalTest instructions: Give the number of N, select some of the number of options for the sum of squares. Training Guide Topics.Analysis: Each of the numbers decomposes factorization. Like 4, 6, 10, 15,,,,,,,,,,,,,,,,,,,,,,,,,,, and if P is squared, then the exponent on each factorization is even, and the X1 factor of 2 is already an even number. Can be converted to XOR 0 to determine the even, that is, an odd number of 1, even set to 0, and then n numbers m factorization augmented matri

, such that 1 = a, 2 = b etc. The value 0 is transcribed to "*" (an asterisk). While transcribing messages, the linguists simply loop from k = 1 to N, and append the character corresponding to the Valu E of F (k) at the end of the string.The backward transcription procedure, has however, turned off to being too complex for the linguists of handle by themselves. You is therefore assigned the task of writing a program this converts a set of strings to their corresponding Extra Terr Estial number s

Title: http://acm.hdu.edu.cn/showproblem.php?pid=4305Test instructions: Compare bare spanning tree count issues.How do I handle spanning tree count issues? Kirchhoff Matrix:If i==j kir[i][j] = the degree of IIf i!=j kir[i][j] = the number of parallel edges of I to J negativeThat is, the Kirchhoff matrix = the degree matrix-adjacency matrixThe Kirchhoff matrix is deleted from row I and column I, and the value of the determinant of the remaining i-1 order is the number of spanning trees. (Proof sl

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

Label: style blog HTTP Io OS AR for SP on Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3915 This topic is Gaussian deyuan linked to game theory. The game involved in this question is the NIM game. The conclusion is: when the first hand is in a strange situation (the number of stones is different from each other or 0), it will be defeated. Here, the final answer is 2 ^ K and K indicates the number of free yuan changes because free yuan can be

Question: In some dependent lights (that is, this is changed), we need to change the status of several lights to reach the target State from the initial state. Each lamp can be changed once at most, total number of solutions. Analysis: Since each lamp can only be changed once, the final state is only related to the change of the lamp related to the lamp, and it is easy to think of setting up a equations to solve the problem. As for solving the equations, Gau