南邮1340 矩阵逆 初等行转换

题目链接; noj1340

好长时间没写博客了， 今天刚学矩阵逆， 找到裸题， 试试， 嘿嘿

题解: 初等行转换， (A, E) -> (E, A^-1);

code：

#include<cstdio>#include<cstring>#include<cmath>#include<vector>using namespace std;const int maxn = 100 + 5;int n, word;vector<double> a[maxn], c[maxn], t[maxn];double  temp;inline vector<double> operator * (vector<double>a, double b) {vector<double> res(n, 0);for (int i = 0; i < n; i++)res[i] = a[i] * b;return res;}inline vector<double> operator - (vector<double>a, vector<double> b) {vector<double> res(n, 0);for (int i = 0; i < n; i++)res[i] = a[i] - b[i];return res;}inline void inverse() {for (int i = 0; i < n; i++)c[i] = vector<double>(n, 0);for (int i = 0; i < n; i++)c[i][i] = 1;for (int i = 0; i < n; i++) {word = 0;for (int j = i; j < n; j++) {if (fabs(a[j][i]) > 0) {word = 1;swap(a[i], a[j]);swap(c[i], c[j]);break;}}//printf("word : %d\n", word);if (!word)return;c[i] = c[i] * (1 / a[i][i]);a[i] = a[i] * (1 / a[i][i]);for (int j = 0; j < n; j++) {if (j != i && fabs(a[j][i]) > 0) {c[j] = c[j] - c[i] * a[j][i];a[j] = a[j] - a[i] * a[j][i];}}}}int main() {//freopen("in.txt", "r", stdin);while (~scanf("%d", &n)) {//printf("%d\n", n);for (int i = 0; i < n; i++){for (int j = 0; j < n; j++) {scanf("%lf", &temp);a[i].push_back(temp);}}inverse();/*for (int i = 0; i < n; i++){//a[i](n, 0);for (int j = 0; j < n; j++)printf("  %lf", c[i][j]);puts("");}*/for (int i = 0; i < n; i++){//a[i](n, 0);for (int j = 0; j < n; j++) {scanf("%lf", &temp);t[i].push_back(temp);}}if (word) {for (int i = 0; i < n; i++){//a[i](n, 0);for (int j = 0; j < n; j++){//printf("%lf, %lf\n", t[i][j], c[i][j]);if (fabs(t[i][j] - c[i][j]) > 1e-6){word = 0;break;}}if (!word)break;}printf("%s\n", word ? "YES" : "NO");}else {printf("NO\n");}}}