2.c

实现 了 数值分析 课程 中得 几个 算法

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#include <stdio.h>
#include <math.h>
#define M 20

int main(int argc, char *argv[])
{
  int n;
  int i, j, k;
  int mi;
  double mx, tmp;
  static double a[M][M], b[M], x[M], y[M], z[M];
  static double L[M][M], d[M];
  printf("\nInput n value(dim of Ax=b):");
  scanf("%d", &n);
  if (n > M)
  {
    printf("The input n is larger then M,please redefine the M.\n");
    return 1;
  }
  if (n <= 0)
  {
    printf("Please input a number between 1 and %d.\n", M);
    return 1;
  }
  printf("now input the marrix a(i,j),i,j=0...,%d:\n", n - 1);
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      scanf("%lf", &a[i][j]);
  printf("now input the marrix b(i,j),i,j=0...,%d:\n", n - 1);
  for (i = 0; i < n; i++)
    scanf("%lf", &b[i]);
  for (i = 0; i < n; i++)
    L[i][j] = 1;
  for (k = 0; k < n; k++)
  {
    d[k] = a[k][k];
    for (j = 0; j <= k - 1; j++)
      d[k] -= (L[k][j] *L[k][j] *d[j]);
    for (i = k + 1; i < n; i++)
    {
      L[i][k] = a[i][k];
      for (j = 0; j <= k - 1; j++)
        L[i][k] -= (L[i][j] *L[k][j] *d[j]);
      L[i][k] /= d[k];
    }
  }


  for (i = 0; i < n; i++)
  {
    z[i] = b[i];
    for (j = 0; j <= i - 1; j++)
      z[i] -= L[i][j] *z[j];
  }
  for (i = 0; i < n; i++)
    y[i] = z[i] / d[i];
  for (i = n - 1; i >= 0; i--)
  {
    x[i] = y[i];
    for (j = i + 1; j < n; j++)
      x[i] -= L[j][i] *x[j];
  }
  printf("Solve...x-i=\n");
  for (i = 0; i < n; i++)
    printf("%f\n", x[i]);
  system("pause");
  return 0;

}