使用Gauss-Jordan 消元法的计算

Z时代
2024-01-10
分类：技术分享

**学校的作业，让程序最后输出成如图这样，代码看了半天不知道怎么搞，自己的代码输出后是第三张图那样，一堆乱数，实在不明白，哪位高手来替小弟解下惑!

图片说明

代码~~~~~~~

#include

#define N 100

void output(int n, int st, double a[100][N], double b[100])

{

int i, j;

printf("\n\nStep  %3d\n", st);
for (i = 1;i <= n;i++) {
for (j = 1;j <= n;j++)
printf("   %8.2f", a[i][j]);
printf("   %8.2f\n", b[i]);
}

}

int main(void)

{

int i, j, k, n, q;

double p;

double a[N][N], b[N], x[N];

printf("namber of demension(n) =");
scanf("%d", &n);
for (i = 1;i <= n;i++) {
for (j = 1;j <= n;j++) {
printf("a[%d][%d] =", i, j);
scanf("%lf", &a[i][j]);
}
printf("b[%d] =", i);
scanf("%lf", &b[i]);
}
printf("\n\nGauss eliminaition method\n\n");
printf("input data\n");
for (j = 1;j <= n;j++)
printf("         x%d", j);
printf("       rhs\n");
for (i = 1;i <= n;i++) {
for (j = 1;j <= n;j++)
printf("   %8.2f", a[i][j]);
printf("   %8.2f\n", b[i]);
}
for (i = 1;i <= n;i++) {
p = a[i][i];
for (j = i;j <= n;j++)
a[i][j] = a[i][j] / p;
b[i] = b[i] / p;
for (k = 1;k <= n;k++) {
if (k != i) {
q = a[k][j];
for (j = i;j <= n;j++)
a[k][j] = a[k][j] - q * a[i][j];
b[k] = b[k] - q * b[i];
}
}
output(n, i, a, b);
}
printf("\n\n solution :\n\n");
for (i = 1; i <= n; i++)
printf(" x[%d] =b[%d] %8.2f¥n", i, x[i], b[i]);
return 0;