Poj 2947 widget factory Gaussian elimination

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 infinite solutions.

In addition, after making this sad question, it seems that there are some problems with Gaussian yuan.

# Include <cstdio> # include <cstring> # include <algorithm> # include <map> # include <set> # include <bitset> # include <queue> # include <stack> # include <string> # include <iostream> # include <cmath> # include <climits> using namespace STD; const int maxn = 400; int A [maxn] [maxn], n, m, ANS [maxn], fans [maxn], anscnt; char T1 [1024], t2 [1024]; int MP (char * s) {If (strcmp (S, "mon") = 0) return 1; else if (strcmp (S, "Tue") = 0) retu Rn 2; else if (strcmp (S, "wed") = 0) return 3; else if (strcmp (S, "Thu") = 0) return 4; else if (strcmp (S, "fri") = 0) return 5; else if (strcmp (S, "sat") = 0) return 6; else return 7;} int gcd (int A, int B) {return B = 0? A: gcd (B, A % B);} int lcm (int A, int B) {return a/gcd (a, B) * B ;} int getmod (int x) {return (X % 7 + 7) % 7;} //-1 indicates no solution, 0 indicates a unique solution, and 1 indicates that multiple solutions int Gauss () {int row, Col; For (ROW = 0, Col = 0; row <M & Col <n; Col ++) {int K = row; while (A [k] [col] = 0 & K <m) K ++; If (k> = m) continue; For (INT I = 0; I <= N; I ++) Swap (A [row] [I], a [k] [I]); For (INT I = 0; I <m; I ++) if (I! = Row & A [I] [col]! = 0) {int lc = lcm (ABS (A [row] [col]), ABS (A [I] [col]); int C1 = LC/A [row] [col], C2 = LC/A [I] [col]; for (Int J = 0; j <= N; j ++) {A [I] [J] = getmod (A [I] [J] * C2-A [row] [J] * C1 );}} row ++;} // unsolvable for (INT I = row; I <m; I ++) if (a [I] [N]) Return-1; // Infinitely multiple solutions if (row <n) return 1; // Unique Solution for (INT I = row-1; I> = 0; I --) {int right = A [I] [N], Left = A [I] [I]; for (Int J = I + 1; j <n; j ++) {right = getmod (right-[ I] [J] * ans [J]);} while (right % left! = 0) Right + = 7; ans [I] = right/left % 7; If (ANS [I] <3) ans [I] + = 7 ;} return 0;} void PA () {for (INT I = 0; I <m; I ++) {for (Int J = 0; j <= N; j ++) {printf ("% d", a [I] [J]);} puts ("") ;}} int main () {// freopen ("in.txt", "r", stdin); While (scanf ("% d", & N, & M )! = EOF) {If (n = 0 & M = 0) break; int K, TMP; anscnt = 0; memset (A, 0, sizeof ()); for (INT I = 0; I <m; I ++) {scanf ("% d % S % s", & K, T1, T2 ); A [I] [N] = getmod (MP (T2)-MP (T1) + 1); For (Int J = 0; j <K; j ++) {scanf ("% d", & TMP); A [I] [TMP-1] = getmod (A [I] [TMP-1] + 1 );}} int ret = Gauss (); If (ret =-1) puts ("inconsistent data. "); else if (ret = 1) puts (" multiple solutions. "); else {for (INT I = 0; I <n; I ++) printf (" % d ", ANS [I]); puts ("") ;}} return 0 ;}

　　

Poj 2947 widget factory Gaussian elimination