计算方法之用高斯列主元消去法求线性方程组

/************************************** 用高斯列主元消去法求线性方程组* * 2*x1 + 2*x2 + 3*x3 = 3*{4*x1 + 7*x2 + 7*x3 = 1* -2*x1+ 4*x2 + 5*x3 = -7***************************************/

#include<stdio.h>#include<math.h>#include<conio.h>#include<string.h>#define N 3int main() {static double a[N][N] = { { 2, 2, 3 }, { 4, 7, 7 }, { -2, 4, 5 } };double b[N] = { 3, 1, -7 };double x[N] = { 0, 0, 0 };double r, s, e;int k, i, j, p, flag = 1;for (k = 0; k < N - 1; k++) {p = k;e = a[k][k];for (i = k; i < N; i++)if (fabs(a[i][k]) > e) {e = fabs(a[i][k]);p = i;}for (j = k; j < N; j++) {s = a[k][j];a[k][j] = a[p][j];a[p][j] = s;}s = b[k];b[k] = b[p];b[p] = s;if (a[k][k] == 0) {printf("Gauss-Method does not run!");flag = 0;break;} else {for (i = k + 1; i < N; i++) {r = a[i][k] / a[k][k];if (a[k][k] != 0) {for (j = k; j < N; j++)a[i][j] = a[i][j] - r * a[k][j];}b[i] -= r * b[k];}}}if (flag) {x[N - 1] = b[N - 1] / a[N - 1][N - 1];for (i = N - 2; i >= 0; i--) {s = b[i];for (j = i + 1; j < N; j++)s -= a[i][j] * x[j];x[i] = s / a[i][i];}for (i = 0; i < N; i++)printf("x[%d] = %10.7f\n", i + 1, x[i]);}return 0;}