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POJ 2947 Widget Factory 高斯消元

利用高斯消元解同余方程组,并且判断是否是唯一解,如果是唯一输出答案。

根据线性代数里面的只是有如果系数矩阵的秩和增广矩阵的秩相同的话,那么有唯一解。

如果系数矩阵的秩小于增广矩阵,那么是无解。如果秩小于变元的个数,那么有无穷多解。

而且通过做了这题悲伤的发现,之前做的高斯消元好像有一些问题。

#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)		return 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表示无解,0表示有唯一解,1表示有多解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++;	}	//无解	for(int i = row; i < m; i++) if(a[i][n]) return -1;	//无穷多解	if(row < n) return 1;	//唯一解	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 - a[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%d", &n, &m) != EOF) {		if(n == 0 && m == 0) break;		int k, tmp;		anscnt = 0;		memset(a, 0, sizeof(a));		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 高斯消元