## 线性代数方程组的求解 原

RapidBird

#include "stdafx.h"
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
#include "LinearEquation.h"
#include "MatrixAlgo.h"

//求解三对角线方程组的追赶法
int atrde(double b[],int n, int m, double d[])
{
int k,j;
double s;
if(m != (3*n-2))
{
printf("err\n");
return(-2);
}
for(k = 0;k <= n-2;k++)
{
j = 3*k;
s = b[j];
if(fabs(s)+1.0 == 1.0)
{
printf("fail\n");
return(0);
}
b[j+1] = b[j+1]/s;
d[k] = d[k]/s;
b[j+3] = b[j+3]-b[j+2]*b[j+1];
d[k+1] = d[k+1]-b[j+2]*d[k];
}
s = b[3*n-3];
if(fabs(s)+1.0 == 1.0)
{
printf("fail\n");
return(0);
}
d[n-1] = d[n-1]/s;
for(k = n-2;k>=0;k--)
d[k] = d[k]-b[3*k+1]*d[k+1];
return(2);
}

//求解托伯利兹方程组的列文逊方法
int atlvs(double t[], int n, double b[], double x[])
{
int i,j,k;
double a,beta,q,c,h,*y,*s;
s = (double *)malloc(n*sizeof(double));
y = (double *)malloc(n*sizeof(double));
a = t[0];
if(fabs(a)+1.0 == 1.0)
{
free(s);
free(y);
printf("fail\n");
return(-1);
}
y[0] = 1.0;
x[0] = b[0]/a;
for(k = 1; k <= n-1; k++)
{
beta = 0.0;
q = 0.0;
for(j = 0; j <= k-1; j++)
{
beta = beta+y[j]*t[j+1];
q = q+x[j]*t[k-j];
}
if(fabs(a)+1.0 == 1.0)
{
free(s);
free(y);
printf("fail\n");
return(-1);
}
c = -beta/a;
s[0] = c*y[k-1];
y[k] = y[k-1];
if(k != 1)
for(i = 1; i <= k-1; i++)
s[i] = y[i-1]+c*y[k-i-1];
a = a+c*beta;
if(fabs(a)+1.0 == 1.0)
{
free(s);
free(y);
printf("fail\n");
return(-1);
}
h = (b[k]-q)/a;
for(i = 0; i <= k-1; i++)
{
x[i] = x[i]+h*s[i];
y[i] = s[i];
}
x[k] = h*y[k];
}
free(s);
free(y);
return(1);
}

//求解对称方程组的分解法
int aldle(double a[], int n, int m, double c[])
{
int i,j,l,k,u,v,w,k1,k2,k3;
double p;
if(fabs(a[0])+1.0 == 1.0)
{
printf("fail\n");
return(-2);
}
for(i = 1; i <= n-1; i++)
{
u = i*n;
a[u] = a[u]/a[0];
}
for(i = 1; i <= n-2; i++)
{
u = i*n+i;
for(j = 1; j <= i; j++)
{
v = i*n+j-1;
l = (j-1)*n+j-1;
a[u] = a[u]-a[v]*a[v]*a[l];
}
p = a[u];
if(fabs(p)+1.0 == 1.0)
{
printf("fail\n");
return(-2);
}
for(k = i+1; k <= n-1; k++)
{
u = k*n+i;
for(j = 1; j <= i; j++)
{
v = k*n+j-1;
l = i*n+j-1;
w = (j-1)*n+j-1;
a[u] = a[u]-a[v]*a[l]*a[w];
}
a[u] = a[u]/p;
}
}
u = n*n-1;
for(j = 1; j <= n-1; j++)
{
v = (n-1)*n+j-1;
w = (j-1)*n+j-1;
a[u] = a[u]-a[v]*a[v]*a[w];
}
p = a[u];
if(fabs(p)+1.0 == 1.0)
{
printf("fail\n");
return(-2);
}
for(j = 0; j <= m-1; j++)
for(i = 1; i <= n-1; i++)
{
u = i*m+j;
for(k = 1; k <= i; k++)
{
v = i*n+k-1;
w = (k-1)*m+j;
c[u] = c[u]-a[v]*c[w];
}
}
for(i = 1; i <= n-1; i++)
{
u = (i-1)*n+i-1;
for(j = i; j <= n-1; j++)
{
v = (i-1)*n+j;
w = j*n+i-1;
a[v] = a[u]*a[w];
}
}
for(j = 0; j <= m-1; j++)
{
u = (n-1)*m+j;
c[u] = c[u]/p;
for(k = 1; k <= n-1; k++)
{
k1 = n-k;
k3 = k1-1;
u = k3*m+j;
for(k2 = k1; k2 <= n-1; k2++)
{
v = k3*n+k2;
w = k2*m+j;
c[u] = c[u]-a[v]*c[w];
}
c[u] = c[u]/a[k3*n+k3];
}
}
return(2);
}

//高斯-赛德尔迭代法
int agsdl(double a[], double b[], int n, double x[], double eps)
{
int i,j,u,v;
double p,t,s,q;
for(i = 0; i <= n-1; i++)
{
u = i*n+i;
p = 0.0;
x[i] = 0.0;
for(j = 0; j <= n-1; j++)
if(i != j)
{
v = i*n+j;
p = p+fabs(a[v]);
}
if(p>=fabs(a[u]))
{
printf("fail\n");
return(-1);
}
}
p = eps+1.0;
while(p>=eps)
{
p = 0.0;
for(i = 0; i <= n-1; i++)
{
t = x[i];
s = 0.0;
for(j = 0; j <= n-1; j++)
if(j != i)
s = s+a[i*n+j]*x[j];
x[i] = (b[i]-s)/a[i*n+i];
q = fabs(x[i]-t)/(1.0+fabs(x[i]));
if(q>p)
p = q;
}
}
return(1);
}

//求解对称正定方程组的共轭梯度法
void agrad(double a[], int n, double b[], double eps, double x[])
{
int i,k;
double *p,*r,*s,*q,alpha,beta,d,e;
p = (double *)malloc(n*sizeof(double));
r = (double *)malloc(n*sizeof(double));
s = (double *)malloc(n*sizeof(double));
q = (double *)malloc(n*sizeof(double));
for(i = 0; i <= n-1; i++)
{
x[i] = 0.0;
p[i] = b[i];
r[i] = b[i];
}
i = 0;
while(i <= n-1)
{
brmul(a,p,n,n,1,s);
d = 0.0;
e = 0.0;
for(k = 0; k <= n-1; k++)
{
d = d+p[k]*b[k];
e = e+p[k]*s[k];
}
alpha = d/e;
for(k = 0; k <= n-1; k++)
x[k] = x[k]+alpha*p[k];
brmul(a,x,n,n,1,q);
d = 0.0;
for(k = 0; k <= n-1; k++)
{
r[k] = b[k]-q[k];
d = d+r[k]*s[k];
}
beta = d/e;
d = 0.0;
for(k = 0; k <= n-1; k++)
d = d+r[k]*r[k];
d = sqrt(d);
if(d<eps)
{
free(p);
free(r);
free(s);
free(q);
return;
}
for(k = 0; k <= n-1; k++)
p[k] = r[k]-beta*p[k];
i = i+1;
}
free(p);
free(r);
free(s);
free(q);
return;
}

//求解线性最小二乘问题的豪斯荷尔德变换法
int agmqr(double a[], int m, int n, double b[], double q[])
{
int i,j;
double d,*c;
c = (double *)malloc(n*sizeof(double));
i = bmaqr(a,m,n,q);
if(i == 0)
{
free(c);
return(0);
}
for(i = 0; i <= n-1; i++)
{
d = 0.0;
for(j = 0; j <= m-1; j++)
d = d+q[j*m+i]*b[j];
c[i] = d;
}
b[n-1] = c[n-1]/a[n*n-1];
for(i = n-2; i>=0; i--)
{
d = 0.0;
for(j = i+1; j <= n-1; j++)
d = d+a[i*n+j]*b[j];
b[i] = (c[i]-d)/a[i*n+i];
}
free(c);
return(1);
}

//求解线性最小二乘问题的广义逆法
int agmiv(double a[], int m, int n, double b[], double x[], double aa[], double eps, double u[], double v[], int ka)
{
int i,j;
i = bginv(a,m,n,aa,eps,u,v,ka);
if(i<0)
return(-1);
for(i = 0; i <= n-1; i++)
{
x[i] = 0.0;
for(j = 0; j <= m-1; j++)
x[i] = x[i]+aa[i*m+j]*b[j];
}
return(1);
}

//全选主元高斯-约当消去法
int agjdn(double a[], double b[], int n, int m)
{
int *js,l,k,i,j,is,p,q;
double d,t;
js = (int *)malloc(n*sizeof(int));
l = 1;
for(k = 0;k <= n-1;k++)
{
d = 0.0;
for(i = k;i <= n-1;i++)
for(j = k;j <= n-1;j++)
{
t = fabs(a[i*n+j]);
if(t>d)
{
d = t;
js[k] = j;
is = i;
}
}
if(d+1.0 == 1.0)
l = 0;
else
{
if(js[k] != k)
for(i = 0;i <= n-1;i++)
{
p = i*n+k;
q = i*n+js[k];
t = a[p];
a[p] = a[q];
a[q] = t;
}
if(is != k)
{
for(j = k;j <= n-1;j++)
{
p = k*n+j;
q = is*n+j;
t = a[p];
a[p] = a[q];
a[q] = t;
}
for(j = 0;j <= m-1;j++)
{
p = k*m+j;
q = is*m+j;
t = b[p];
b[p] = b[q];
b[q] = t;
}
}
}
if(l == 0)
{
free(js);
printf("fail\n");
return(0);
}
d = a[k*n+k];
for(j = k+1;j <= n-1;j++)
{
p = k*n+j;
a[p] = a[p]/d;
}
for(j = 0;j <= m-1;j++)
{
p = k*m+j;
b[p] = b[p]/d;
}
for(j = k+1;j <= n-1;j++)
for(i = 0;i <= n-1;i++)
{
p = i*n+j;
if(i != k)
a[p] = a[p]-a[i*n+k]*a[k*n+j];
}
for(j = 0;j <= m-1;j++)
for(i = 0;i <= n-1;i++)
{
p = i*m+j;
if(i != k)
b[p] = b[p]-a[i*n+k]*b[k*m+j];
}
}
for(k = n-1;k>=0;k--)
if(js[k] != k)
for(j = 0;j <= m-1;j++)
{
p = k*m+j;
q = js[k]*m+j;
t = b[p];
b[p] = b[q];
b[q] = t;
}
free(js);
return(1);
}

//求解大型稀疏方程组的全选主元高斯-约当消去法
int aggje(double a[], int n, double b[])
{
int *js,i,j,k,is,u,v;
double d,t;
js = (int *)malloc(n*sizeof(int));
for(k = 0; k <= n-1; k++)
{
d = 0.0;
for(i = k; i <= n-1; i++)
for(j = k; j <= n-1; j++)
{
t = fabs(a[i*n+j]);
if(t>d)
{
d = t;
js[k] = j;
is = i;
}
}
if(d+1.0 == 1.0)
{
free(js);
printf("fail\n");
return(0);
}
if(is != k)
{
for(j = k; j <= n-1; j++)
{
u = k*n+j;
v = is*n+j;
t = a[u];
a[u] = a[v];
a[v] = t;
}
t = b[k];
b[k] = b[is];
b[is] = t;
}
if(js[k] != k)
for(i=0; i <= n-1; i++)
{
u = i*n+k;
v = i*n+js[k];
t = a[u];
a[u] = a[v];
a[v] = t;
}
t = a[k*n+k];
for(j = k+1; j <= n-1; j++)
{
u = k*n+j;
if(a[u] != 0.0)
a[u] = a[u]/t;
}
b[k] = b[k]/t;
for(j = k+1; j <= n-1; j++)
{
u = k*n+j;
if(a[u] != 0.0)
{
for(i = 0; i <= n-1; i++)
{
v = i*n+k;
if((i != k)&&(a[v] != 0.0))
{
is = i*n+j;
a[is] = a[is]-a[v]*a[u];
}
}
}
}
for(i = 0; i <= n-1; i++)
{
u = i*n+k;
if((i != k)&&(a[u] != 0.0))
b[i] = b[i]-a[u]*b[k];
}
}
for(k = n-1; k>=0; k--)
if(k != js[k])
{
t = b[k];
b[k] = b[js[k]];
b[js[k]] = t;
}
free(js);
return(1);
}

//全选主元高斯消去法
int agaus(double a[], double b[], int n)
{
int *js,l,k,i,j,is,p,q;
double d,t;
js = (int *)malloc(n*sizeof(int));
l = 1;
for(k = 0;k <= n-2;k++)
{
d = 0.0;
for(i = k;i <= n-1;i++)
for(j = k;j <= n-1;j++)
{
t = fabs(a[i*n+j]);
if(t>d)
{
d = t;
js[k] = j;
is = i;
}
}
if(d+1.0 == 1.0)
l = 0;
else
{
if(js[k] != k)
for(i = 0;i <= n-1;i++)
{
p = i*n+k;
q = i*n+js[k];
t = a[p];
a[p] = a[q];
a[q] = t;
}
if(is != k)
{
for(j = k;j <= n-1;j++)
{
p = k*n+j;
q = is*n+j;
t = a[p];
a[p] = a[q];
a[q] = t;
}
t = b[k];
b[k] = b[is];
b[is] = t;
}
}
if(l == 0)
{
free(js);
printf("fail\n");
return(0);
}
d = a[k*n+k];
for(j = k+1;j <= n-1;j++)
{
p = k*n+j;
a[p] = a[p]/d;
}
b[k] = b[k]/d;
for(i = k+1;i <= n-1;i++)
{
for(j = k+1;j <= n-1;j++)
{
p = i*n+j;
a[p] = a[p]-a[i*n+k]*a[k*n+j];
}
b[i] = b[i]-a[i*n+k]*b[k];
}
}
d = a[(n-1)*n+n-1];
if(fabs(d)+1.0 == 1.0)
{
free(js);
printf("fail\n");
return(0);
}
b[n-1] = b[n-1]/d;
for(i = n-2;i>=0;i--)
{
t = 0.0;
for(j = i+1;j <= n-1;j++)
t = t+a[i*n+j]*b[j];
b[i] = b[i]-t;
}
js[n-1] = n-1;
for(k = n-1;k>=0;k--)
if(js[k] != k)
{
t = b[k];
b[k] = b[js[k]];
b[js[k]] = t;
}
free(js);
return(1);
}

//复系数方程组的全选主元高斯-约当消去法
int acjdn(double ar[], double ai[], double br[], double bi[], int n, int m)
{
int *js,l,k,i,j,is,u,v;
double p,q,s,d;
js = (int *)malloc(n*sizeof(int));
for(k = 0;k <= n-1;k++)
{
d = 0.0;
for(i = k;i <= n-1;i++)
for(j = k;j <= n-1;j++)
{
u = i*n+j;
p = ar[u]*ar[u]+ai[u]*ai[u];
if(p>d)
{
d = p;
js[k] = j;
is = i;
}
}
if(d+1.0 == 1.0)
{
free(js);
printf("err**fail\n");
return(0);
}
if(is != k)
{
for(j = k;j <= n-1;j++)
{
u = k*n+j;
v = is*n+j;
p = ar[u];
ar[u] = ar[v];
ar[v] = p;
p = ai[u];
ai[u] = ai[v];
ai[v] = p;
}
for(j = 0;j <= m-1;j++)
{
u = k*m+j;
v = is*m+j;
p = br[u];
br[u] = br[v];
br[v] = p;
p = bi[u];
bi[u] = bi[v];
bi[v] = p;
}
}
if(js[k] != k)
for(i = 0;i <= n-1;i++)
{
u = i*n+k;
v = i*n+js[k];
p = ar[u];
ar[u] = ar[v];
ar[v] = p;
p = ai[u];
ai[u] = ai[v];
ai[v] = p;
}
v = k*n+k;
for(j = k+1;j <= n-1;j++)
{
u = k*n+j;
p = ar[u]*ar[v];
q = -ai[u]*ai[v];
s = (ar[v]-ai[v])*(ar[u]+ai[u]);
ar[u] = (p-q)/d;
ai[u] = (s-p-q)/d;
}
for(j = 0;j <= m-1;j++)
{
u = k*m+j;
p = br[u]*ar[v];
q = -bi[u]*ai[v];
s = (ar[v]-ai[v])*(br[u]+bi[u]);
br[u] = (p-q)/d;
bi[u] = (s-p-q)/d;
}
for(i = 0;i <= n-1;i++)
if(i != k)
{
u = i*n+k;
for(j = k+1;j <= n-1;j++)
{
v = k*n+j;
l = i*n+j;
p = ar[u]*ar[v];
q = ai[u]*ai[v];
s = (ar[u]+ai[u])*(ar[v]+ai[v]);
ar[l] = ar[l]-p+q;
ai[l] = ai[l]-s+p+q;
}
for(j = 0;j <= m-1;j++)
{
l = i*m+j;
v = k*m+j;
p = ar[u]*br[v];
q = ai[u]*bi[v];
s = (ar[u]+ai[u])*(br[v]+bi[v]);
br[l] = br[l]-p+q;
bi[l] = bi[l]-s+p+q;
}
}
}
for(k = n-1;k>=0;k--)
if(js[k] != k)
for(j = 0;j <= m-1;j++)
{
u = k*m+j;
v = js[k]*m+j;
p = br[u];
br[u] = br[v];
br[v] = p;
p = bi[u];
bi[u] = bi[v];
bi[v] = p;
}
free(js);
return(1);
}

//求解对称正定方程组的平方根法
int achol(double a[], int n, int m, double d[])
{
int i,j,k,u,v;
if((a[0]+1.0 == 1.0)||(a[0]<0.0))
{
printf("fail\n");
return(-2);
}
a[0] = sqrt(a[0]);
for(j = 1; j <= n-1; j++)
a[j] = a[j]/a[0];
for(i = 1; i <= n-1; i++)
{
u = i*n+i;
for(j = 1; j <= i; j++)
{
v = (j-1)*n+i;
a[u] = a[u]-a[v]*a[v];
}
if((a[u]+1.0 == 1.0)||(a[u]<0.0))
{
printf("fail\n");
return(-2);
}
a[u] = sqrt(a[u]);
if(i != (n-1))
{
for(j = i+1; j <= n-1; j++)
{
v = i*n+j;
for(k = 1; k <= i; k++)
a[v] = a[v]-a[(k-1)*n+i]*a[(k-1)*n+j];
a[v] = a[v]/a[u];
}
}
}
for(j = 0; j <= m-1; j++)
{
d[j] = d[j]/a[0];
for(i = 1; i <= n-1; i++)
{
u = i*n+i;
v = i*m+j;
for(k = 1; k <= i; k++)
d[v] = d[v]-a[(k-1)*n+i]*d[(k-1)*m+j];
d[v] = d[v]/a[u];
}
}
for(j = 0; j <= m-1; j++)
{
u = (n-1)*m+j;
d[u] = d[u]/a[n*n-1];
for(k = n-1; k>=1; k--)
{
u = (k-1)*m+j;
for(i = k; i <= n-1; i++)
{
v = (k-1)*n+i;
d[u] = d[u]-a[v]*d[i*m+j];
}
v = (k-1)*n+k-1;
d[u] = d[u]/a[v];
}
}
return(2);
}

//复系数方程组的全选主元高斯消去法
int acgas(double ar[], double ai[], int n, double br[], double bi[])
{
int *js,l,k,i,j,is,u,v;
double p,q,s,d;
js = (int *)malloc(n*sizeof(int));
for(k = 0;k <= n-2;k++)
{
d = 0.0;
for(i = k;i <= n-1;i++)
for(j = k;j <= n-1;j++)
{
u = i*n+j;
p = ar[u]*ar[u]+ai[u]*ai[u];
if(p>d)
{
d = p;
js[k] = j;
is = i;
}
}
if(d+1.0 == 1.0)
{
free(js);
printf("err**fail\n");
return(0);
}
if(is != k)
{
for(j = k;j <= n-1;j++)
{
u = k*n+j;
v = is*n+j;
p = ar[u];
ar[u] = ar[v];
ar[v] = p;
p = ai[u];
ai[u] = ai[v];
ai[v] = p;
}
p = br[k];
br[k] = br[is];
br[is] = p;
p = bi[k];
bi[k] = bi[is];
bi[is] = p;
}
if(js[k] != k)
for(i = 0;i <= n-1;i++)
{
u = i*n+k;
v = i*n+js[k];
p = ar[u];
ar[u] = ar[v];
ar[v] = p;
p = ai[u];
ai[u] = ai[v];
ai[v] = p;
}
v = k*n+k;
for(j = k+1;j <= n-1;j++)
{
u = k*n+j;
p = ar[u]*ar[v];
q = -ai[u]*ai[v];
s = (ar[v]-ai[v])*(ar[u]+ai[u]);
ar[u] = (p-q)/d;
ai[u] = (s-p-q)/d;
}
p = br[k]*ar[v];
q = -bi[k]*ai[v];
s = (ar[v]-ai[v])*(br[k]+bi[k]);
br[k] = (p-q)/d;
bi[k] = (s-p-q)/d;
for(i = k+1;i <= n-1;i++)
{
u = i*n+k;
for(j = k+1;j <= n-1;j++)
{
v = k*n+j;
l = i*n+j;
p = ar[u]*ar[v];
q = ai[u]*ai[v];
s = (ar[u]+ai[u])*(ar[v]+ai[v]);
ar[l] = ar[l]-p+q;
ai[l] = ai[l]-s+p+q;
}
p = ar[u]*br[k];
q = ai[u]*bi[k];
s = (ar[u]+ai[u])*(br[k]+bi[k]);
br[i] = br[i]-p+q;
bi[i] = bi[i]-s+p+q;
}
}
u = (n-1)*n+n-1;
d = ar[u]*ar[u]+ai[u]*ai[u];
if(d+1.0 == 1.0)
{
free(js);
printf("err**fail\n");
return(0);
}
p = ar[u]*br[n-1];
q = -ai[u]*bi[n-1];
s = (ar[u]-ai[u])*(br[n-1]+bi[n-1]);
br[n-1] = (p-q)/d;
bi[n-1] = (s-p-q)/d;
for(i = n-2;i>=0;i--)
for(j = i+1;j <= n-1;j++)
{
u = i*n+j;
p = ar[u]*br[j];
q = ai[u]*bi[j];
s = (ar[u]+ai[u])*(br[j]+bi[j]);
br[i] = br[i]-p+q;
bi[i] = bi[i]-s+p+q;
}
js[n-1] = n-1;
for(k = n-1;k>=0;k--)
if(js[k] != k)
{
p = br[k];
br[k] = br[js[k]];
br[js[k]] = p;
p = bi[k];
bi[k] = bi[js[k]];
bi[js[k]] = p;
}
free(js);
return(1);
}

//病态方程组的求解
int abint(double a[], int n, double b[], double eps, double x[])
{
int i,j,k,kk;
double *p,*r,*e,q,qq;
p = (double *)malloc(n*n*sizeof(double));
r = (double *)malloc(n*sizeof(double));
e = (double *)malloc(n*sizeof(double));
i = 60;
for(k = 0; k <= n-1; k++)
for(j = 0; j <= n-1; j++)
p[k*n+j] = a[k*n+j];
for(k = 0; k <= n-1; k++)
x[k] = b[k];
kk = agaus(p,x,n);
if(kk == 0)
{
free(p);
free(r);
free(e);
return(kk);
}
q = 1.0+eps;
while(q>=eps)
{
if(i == 0)
{
free(p);
free(r);
free(e);
return(i);
}
i = i-1;
brmul(a,x,n,n,1,e);
for( k = 0; k <= n-1; k++)
r[k] = b[k]-e[k];
for( k = 0; k <= n-1; k++)
for( j = 0; j <= n-1; j++)
p[k*n+j] = a[k*n+j];
kk = agaus(p,r,n);
if(kk == 0)
{
free(p);
free(r);
free(e);
return(kk);
}
q = 0.0;
for( k = 0; k <= n-1; k++)
{
qq = fabs(r[k])/(1.0+fabs(x[k]+r[k]));
if(qq>q)
q = qq;
}
for( k = 0; k <= n-1; k++)
x[k] = x[k]+r[k];
}
free(p);
free(r);
free(e);
return(1);
}

//一般带型方程组的求解
int aband(double b[], double d[], int n, int l, int il, int m)
{
int ls,k,i,j,is,u,v;
double p,t;
if(il != (2*l+1))
{
printf("fail\n");
return(-2);
}
ls = l;
for(k = 0;k <= n-2;k++)
{
p = 0.0;
for(i = k;i <= ls;i++)
{
t = fabs(b[i*il]);
if(t>p)
{
p = t;
is = i;
}
}
if(p+1.0 == 1.0)
{
printf("fail\n");
return(0);
}
for(j = 0;j <= m-1;j++)
{
u = k*m+j;
v = is*m+j;
t = d[u];
d[u] = d[v];
d[v] = t;
}
for(j = 0;j <= il-1;j++)
{
u = k*il+j;
v = is*il+j;
t = b[u];
b[u] = b[v];
b[v] = t;
}
for(j = 0;j <= m-1;j++)
{
u = k*m+j;
d[u] = d[u]/b[k*il];
}
for(j = 1;j <= il-1;j++)
{
u = k*il+j;
b[u] = b[u]/b[k*il];
}
for(i = k+1;i <= ls;i++)
{
t = b[i*il];
for(j = 0;j <= m-1;j++)
{
u = i*m+j;
v = k*m+j;
d[u] = d[u]-t*d[v];
}
for(j = 1;j <= il-1;j++)
{
u = i*il+j;
v = k*il+j;
b[u-1] = b[u]-t*b[v];
}
u = i*il+il-1;
b[u] = 0.0;
}
if(ls != (n-1))
ls = ls+1;
}
p = b[(n-1)*il];
if(fabs(p)+1.0 == 1.0)
{
printf("fail\n");
return(0);
}
for(j = 0;j <= m-1;j++)
{
u = (n-1)*m+j;
d[u] = d[u]/p;
}
ls = 1;
for(i = n-2;i>=0;i--)
{
for(k = 0;k <= m-1;k++)
{
u = i*m+k;
for(j = 1;j <= ls;j++)
{
v = i*il+j;
is = (i+j)*m+k;
d[u] = d[u]-b[v]*d[is];
}
}
if(ls != (il-1))
ls = ls+1;
}
return(2);
}

----根据《C语言常用算法程序集》整理

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