// 解线性方程。data[count*(count+1)]矩阵数组;count:方程元数;
// Answer[count]:求解数组 。返回:0求解成功,-1无解或者无穷解
int LinearEquations(double *data, int count, double *Answer)
{
int j, m, n;
double tmp, **dat, *d = data;
dat = (double**)malloc(count * sizeof(double*));
for (m = 0; m < count; m ++, d += (count + 1))
{
dat[m] = (double*)malloc((count + 1) * sizeof(double));
memcpy(dat[m], d, (count + 1) * sizeof(double));
}
d = (double*)malloc((count + 1) * sizeof(double));
for (m = 0; m < count - 1; m ++)
{
// 如果主对角线元素为0,行交换
for (n = m + 1; n < count && dat[m][m] == 0.0; n ++)
{
if ( dat[n][m] != 0.0)
{
memcpy(d, dat[m], (count + 1) * sizeof(double));
memcpy(dat[m], dat[n], (count + 1) * sizeof(double));
memcpy(dat[n], d, (count + 1) * sizeof(double));
}
}
// 行交换后,主对角线元素仍然为0,无解,返回-1
if (dat[m][m] == 0.0)
{
FreeData(dat, d, count);
return -1;
}
// 消元
for (n = m + 1; n < count; n ++)
{
tmp = dat[n][m] / dat[m][m];
for (j = m; j <= count; j ++)
dat[n][j] -= tmp * dat[m][j];
}
}
for (j = 0; j < count; j ++)
d[j] = 0.0;
// 求得count - 1的元
Answer[count - 1] = dat[count - 1][count] / dat[count - 1][count - 1];
// 逐行代入求各元
for (m = count - 2; m >= 0; m --)
{
for (j = count - 1; j > m; j --)
d[m] += Answer[j] * dat[m][j];
Answer[m] = (dat[m][count] - d[m]) / dat[m][m];
}
FreeData(dat, d, count);
return 0;
}
// Answer[count]:求解数组 。返回:0求解成功,-1无解或者无穷解
int LinearEquations(double *data, int count, double *Answer)
{
int j, m, n;
double tmp, **dat, *d = data;
dat = (double**)malloc(count * sizeof(double*));
for (m = 0; m < count; m ++, d += (count + 1))
{
dat[m] = (double*)malloc((count + 1) * sizeof(double));
memcpy(dat[m], d, (count + 1) * sizeof(double));
}
d = (double*)malloc((count + 1) * sizeof(double));
for (m = 0; m < count - 1; m ++)
{
// 如果主对角线元素为0,行交换
for (n = m + 1; n < count && dat[m][m] == 0.0; n ++)
{
if ( dat[n][m] != 0.0)
{
memcpy(d, dat[m], (count + 1) * sizeof(double));
memcpy(dat[m], dat[n], (count + 1) * sizeof(double));
memcpy(dat[n], d, (count + 1) * sizeof(double));
}
}
// 行交换后,主对角线元素仍然为0,无解,返回-1
if (dat[m][m] == 0.0)
{
FreeData(dat, d, count);
return -1;
}
// 消元
for (n = m + 1; n < count; n ++)
{
tmp = dat[n][m] / dat[m][m];
for (j = m; j <= count; j ++)
dat[n][j] -= tmp * dat[m][j];
}
}
for (j = 0; j < count; j ++)
d[j] = 0.0;
// 求得count - 1的元
Answer[count - 1] = dat[count - 1][count] / dat[count - 1][count - 1];
// 逐行代入求各元
for (m = count - 2; m >= 0; m --)
{
for (j = count - 1; j > m; j --)
d[m] += Answer[j] * dat[m][j];
Answer[m] = (dat[m][count] - d[m]) / dat[m][m];
}
FreeData(dat, d, count);
return 0;
}
dat = (double**)malloc(count * sizeof(double*));
for (m = 0; m < count; m ++, d += (count + 1))
{
dat[m] = (double*)malloc((count + 1) * sizeof(double));
memcpy(dat[m], d, (count + 1) * sizeof(double));
}
这一句是 怎么解释呢