计算方法之用雅克比法求线性方程组

/*************************************
* 用雅克比法求线性方程组
* 
*  5*x1 + 2*x2 + 1*x3 = -12
*{-1*x1 + 4*x2 + 2*x3 = 20
*  2*x1 - 3*x2 +10*x3 = 3
*
**************************************/
#include<stdio.h>
#include<math.h>
#include<conio.h>

#define N 3
#define kmax 100
#define eps 1e-5

static double aa[N][N] = { { 5, 2, 1 }, { -1, 4, 2 }, { 2, -3, 10 } };
static double bb[N] = { -12, 20, 3 };

int main() {
	int k, i, j;
	double d, sum, s, norm;
	double a[N + 1][N + 1], b[N + 1], x[N + 1], y[N + 1];

	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++)
			a[i][j] = aa[i - 1][j - 1];
		b[i] = bb[i - 1];
	}

	for (i = 1; i <= N; i++) {
		x[i] = 0;
	}
	k = 0;

	do {
		k++;
		if (k > kmax) {
			printf("\nThe iterate failed !\n");
			break;
		}
		norm = 0.0;
		for (i = 1; i <= N; i++) {
			sum = 0.0;
			for (j = 1; j <= N; j++) {
				if (j != i)
					sum += a[i][j] * x[j];
			}

			y[i] = (b[i] - sum) / a[i][i];
			d = fabs(y[i] - x[i]);
			if (norm < d)
				norm = d;
		}
		for (i = 1; i <= N; i++)
			x[i] = y[i];
	} while (norm >= eps);
	if (norm < eps) {
		printf("\nThe result is :\n");
		printf("\nk = %d \n", k);
		for (i = 1; i <= N; i++)
			printf("x[%d] = %f\n", i, x[i]);
	}
	return 0;
}