gaussian elimination calculator

The Gaussian element is used to describe the homogeneous equations and determine whether it is the unique solution. If it is the only output answer. In linear algebra, there is only a unique solution if the rank of the coefficient matrix is the same as that of the augmented matrix. If the rank of the coefficient matrix is smaller than the Augmented Matrix, there is no solution. If the rank is less than the number of variable elements, there are infini

At first thought it was a tree DP, and then look at the data range is the rhythm of the matter. The does not want to start in the Gaussian elimination, think binary search can engage, but not how to achieve. Gaussian elimination, we can find that the number of adjacent lamp switch XOR must be 1 (to open), so we can s

- { - 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

[i][k];for (j = col+1; J A[i][col] = 0;}}return 1;}int main (){int I,j,ncas=1;int t;Double a[110][110];scanf ("%d", t);while (t--){memset (A,0.0,sizeof (a));memset (x,0.0,sizeof (x));for (int i=0;imemset (go,-1,sizeof (go));equ=var=100;int all;scanf ("%d", all);while (all--){int t1,t2;scanf ("%d%d", t1,t2);t1--;t2--;Go[t1]=t2;}for (int i=0; i{if (go[i]==-1){if (i{a[i][i]=6.0;a[i][i+1]=-1.0;a[i][i+2]=-1.0;a[i][i+3]=-1.0;a[i][i+4]=-1.0;a[i][i+5]=-1.0;a[i][i+6]=-1.0;x[i]=6;}else if (i==99){a[i][i]=

- intR//Line - intC//column the intI, J, K; -c =0; - for(r=0; rvar; R++,c++){ -MAXR =R; + for(i=r+1; i){ - if(ABS (A[i][c]) >ABS (A[maxr][c])) +MAXR =i; A } at if(MAXR!=R) {//Swap Lines - for(i=0; ivar+1; i++) - swap (a[maxr][i],a[r][i]); - } - for(j=r+1; j){ - if(A[j][c]) { in for(I=c; ivar+1; i++){ -A[j][i] = a[j][i]^A[r][i]; to } + } - } the

1 Output2 1ExercisesGaussian elimination template problem, the specific process see: Http://jingyan.baidu.com/album/39810a23e40c80b636fda63a.html?picindex=1Code:#include #include#include#include#include#include#include#includestring>#includeusing namespacestd;Doublemap[ the][ the],ans[ the];intN;intMain () {//freopen ("a.in", "R", stdin);scanf"%d",N); for(intI=1; i) for(intj=1; j1; j + +) scanf ("%LF",Map[i][j]); for(intI=1; i) { BOOLfl

feet output" Earth ". If there are multiple solutions to the input data, the output "cannot determine". All output contains no quotation marks, please note that the output is case-insensitive. Sample Input 3 5011 1110 1101 0111 1010 1 Sample Output 4Earth? y7m#Earth HINT For 20% of the data, satisfy n=m≤20 N = m≤20 n=m\le20;For 40% of the data, satisfy n=m≤500 N = m≤500 n=m\le500;For 70% of the data, meet n≤500,m≤1000 n≤500, m≤1000 n\le500,m\le1000;For 100% of the data, meet n≤1000,m≤2000 n≤100

Gaussian elimination of the naked question, save as a template bar. It is worth noting that the accuracy of 1e-8, 1e-10, but 1e-6 can pass, it is said that each time to find the largest row elimination can reduce the error. #include

Test instructionsTells you a k+1 point on a sphere of k-dimensional spheres asking the spherical coordinates.SolAt first glance I thought it was a k-dimensional answer and judged the distance ... What a jerk. You look at the messy topics and then forget the most useful basic calculations ...We can see that, assuming the center O, according to the formula that he tells us we can get an equation of any two points and a center point, the equation has K unknowns, then we randomly construct K equatio

finishing his mission is impossible output "impossible! "(No quotes) instead. Inputthere is an integer T (t Outputfor each possible scenario, output a floating number with 2 digits after decimal point If finishing his mission is impossible output one line "impossible! " (No quotes) instead. Sample input2 4 2 0 1 0 50 4 1 0 2 100 Sample output8.14 2.00 Source2012 ACM/ICPC Asia Regional Hangzhou online Recommendliuyiding Converts N points into 2 * N-2 points. Then the

Gaussian elimination:is actually to use matrix elementary transformation to solve the linear equations, but he asked to select each time The main element must be the maximum value. Template#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Gaussian elimination template

nature: For any number of a, a ^ A = 0, we can know that when and only when rank Solution: It is said that it is Gaussian deyuan. If the matrix is converted into a trapezoid, empty rows exist, that is, rank PS: As mentioned above, an exclusive or binary is equivalent to addition or subtraction. There is no conflict with the primary changes of the matrix. Code: 1 /************************************************************************* 2

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents by --- cxlove The painful memory is that some questions in the 4th HDU warm-up match soon had an algorithm to complete the code. It took three hours to debug the question during the competition, and more than two hours to debug the question against official data at night. Finally, I found the problem. It is also a classic switch problem. Due to the recent impact, it is necessary to enumerate free y

][var]; for(intl=idx+1; lvar; l++) if(A[j][l]) X[idx]^=X[l]; CNT+=X[idx]; } ans=min (ans,cnt); } returnans; }}Charstr[ -][ -];intMain () {intT; scanf ("%d",t); GetChar (); while(t--) {n=4; for(intI=0; i4; i++) scanf ("%s", Str[i]); Init (); for(inti =0; I 4; i++) for(intj =0; J 4; J + +) { if(Str[i][j] = ='b') a[i*4+j][ -] =0; Elsea[i*4+j][ -] =1; } intANS1 =solve (); Init (); for(inti =0; I 4; i++)

) { - for(intj = col; J var+1; J + +) - swap (A[k][j], a[max_r][j]); in } - for(inti = k +1; i ) { to if(A[i][col]! =0) { + for(intj = col; J var+1; J + +) { -A[I][J] ^=A[k][j]; the } * } $ }Panax Notoginseng } - for(inti = k; i ) { the if(A[i][col]! =0) + return-1; A } the if(K var)return var-K; + for(inti =var-1; I >=0; i--) { -X[i] = a[i][var]; $ fo

, indicating the number of heaps. Then n integer AI follows, indicating the number of each heap. [Technical Specification] 1. 1 2. 1 3. 1 Outputfor each test case, output "yes" if the second player can win by moving some (can be zero) heaps out, otherwise "no ". Sample Input 31232 2 251 2 3 4 5 Sample output NoYesYesHintFor the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0. Question: RT Idea: think of each

HDU 4975 A simple Gaussian elimination problem. Network Flow + dp on the matrix, hdugaussian Random output is safe. In the same way as hdu4888, it is feasible to run a network stream and then dp. = N ^ 3 dp cannot pass, so change n to 200. = Because the author did not place the problem in a matrix other than 200*200. #include

This question is given as follows: F (1) = C1 (MOD) PF (2) = c2 (MOD) PF (3) = C3 (MOD) p... F (1) = a () * X1 + a (1, 2) * X2 + a (1, 3) * x3... Where a (I, j) = I ^ (J-1 ). In the face of such an equation, our approach is to ignore the (MOD) P on the right side of the equation, because f (I) and CI are the same, therefore, multiplying a number at both ends of an equation does not affect the performance of a given equation. So the rest is the Gaussian