C语言四种方法解线性方程的
2007 编写的。
发了好几天编了个解线性方程组的小程序，可第一次实战就大败而归。经过半天的调试，仍找不出纠正的方法。因为并不是算法的问题，而是因为自己对编译器处理 浮点函数的方法不是很理解。明明D=0的方阵解出来不等于0了，跟踪调试发现，计算过程程序对数据进行了舍去处理，导致最终结果不对。不过如果没有浮点型 的话，这个程序应该算不错了 。不过还是可以方便，大二那些线性代数学习者。
#include<stdio.h>
#include<math.h>
#include<mem.h>
#define NUM 100
void print(void) /* 使用说明 */
{ clrscr();
printf("\n\n\n\n\n\t\t\t\t Introduction \n");
printf("\t*--------------------------------------------------------------*\n");
printf("\t* This program was design for compute linear equations. *\n");
printf("\t* The way of use it is very simple. *\n");
printf("\t* First : Input the number of the equation;(Input 0 to exit) *\n");
printf("\t* Second: Input the coefficient of every eqution; *\n");
printf("\t* Third : Input the constant of every eqution; *\n");
printf("\t* Last : Chose the way you want use to solve the equtions; *\n");
printf("\t* That's all, input any key to run it . . . *\n");
printf("\t*-------------------------By__TJX------------------------------*\n");
getch(); }
void chose(void) /*选择计算方法*/
{ clrscr();
fflush(stdin);
printf("\n\n\n\n\n\t\t**********Introduction********** \n");
printf("\t\t* Chose the way,please. * \n");
printf("\t\t* a : Gauss eliminant. * \n");
printf("\t\t* b : Gauss_yd eliminant. * \n");
printf("\t\t* c : Iterative way. * \n");
printf("\t\t* d : Cramer way. * \n");
printf("\t\t* e : exit. * \n");
printf("\t\t*************By__TJX************ \n");
printf("\t\tPlease choose number :\n");}
void input(double **a1,double b1[],int num) /*数据输入*/
{ int i,j,t;
double *p;
char de1,de2;
do{
printf("Please input array a[%d][%d]: \n",num,num);
printf("Warn: The first number of the array mustn't contain zero! \n");
for(i=1;i<=num;i++)
{printf("Please input array a[%d][]: \n",i);
for(j=1;j<=num;j++)
{t=0;
if(i==1&&j==1)
{ do{
if(t==0) { scanf("%lf",&a1[i][j]); t++;}
else {printf("The input is invalid,please input again:\n"); scanf("%f",&a1[i][j]);}
}while(a1[i][j]==0);}
else scanf("%lf",&a1[i][j]);}}
printf(" \nPlease check the value of array a[%d][%d],press Y to input again.\n",num,num);
do{
de1=getch();
}while(de1!='y'&&de1!='Y'&&de1!='n'&&de1!='N');
}while(de1=='y'||de1=='Y');
do{
printf("Please input array b[%d]: \n",num);
p=b1+1;
for(i=1;i<=num;i++)
scanf("%lf",p++);
printf(" \nPlease check the value of array b[%d],press Y to input again.\n",num);
do{
de2=getch();
}while(de2!='y'&&de2!='Y'&&de2!='n'&&de2!='N');
}while(de2=='y'||de2=='Y');}
int max(double *t1, double x1[],int n) /*迭代子函数*/
{ int i,temp=0;
for(i=1;i<=n;i++)
if(fabs(x1[i]-t1[i])>1e-2) {temp=1;break;}
/* printf(" %d ",temp); */
return temp;
}
int ddcompute(double **a1,double b1[],double x1[],int n) /*迭代法计算*/
{double *t;
int i,j,k=0;
double sum1=0.0,sum2=0.0;
t=(double*)malloc(n*sizeof(double));
printf("\nPlease Input The Initial Value of x:\n");
for(i=1;i<=n;i++)
scanf("%lf",&x1[i]);
do{ k++;
for(i=1;i<=n;i++)
t[i]=x1[i];
for(i=1;i<=n;i++)
{ sum1=0.0;sum2=0.0;
for(j=1;j<=i-1;j++) sum1=sum1+a1[i][j]*x1[j]; /*printf(" sum1= %0.4f ",sum1);*/
for(j=i+1;j<=n;j++) sum2=sum2+a1[i][j]*t[j]; /* printf(" sum2= %0.4f ",sum2);}*/
if(a1[i][i]==0||fabs(sum1)>1e+12||fabs(sum2)>1e+12)
{printf(" \nWarning: These equtions can't be solve by this way!\n Press any Key to continue...");
getch();
free(t);
return 0;}
x1[i]=(b1[i]-sum1-sum2)/a1[i][i];}
}while(max(t,x1,n));
/* for(i=1;i<=n;i++)
{if(i%3==0) printf("\n");
printf(" %.4f ",x1[i]);}*/
free(t);
return 1; }
int gscompute(double **a1,double b1[],double x1[],int n) /*高斯消元法计算*/
{int i,j,k;
double m,sum;
for(k=1;k<=n-1;k++)
for(i=k+1;i<=n;i++)
{ if(a1[k][k]==0) {printf(" \nThese equtions can't be solve is this way.\n Press any Key to continue...");
getch();
return 0; }
if((m=0-a1[i][k]/a1[k][k])==0) {i++; continue;}
else {for(j=k+1;j<=n;j++)
a1[i][j]=a1[i][j]+a1[k][j]*m;
b1[i]=b1[i]+b1[k]*m;}}
/* yi xia ji suan x zhi */
x1[n]=b1[n]/a1[n][n];
for(i=n-1;i>=1;i--)
{sum=0.0;
for(j=n;j>=i+1;j--)
sum=sum+a1[i][j]*x1[j];
x1[i]=(b1[i]-sum)/a1[i][i];}
return 1; }
int gs_ydcompute(double **a1,double b1[],double x1[],int n) /*高斯_约当法计算*/
{int i,j,k;
double m,sum;
for(k=1;k<=n;k++)
{i=1;
while(i<=n)
{ if(a1[k][k]==0) {printf(" \nThese equtions can't be solve is this way.\n Press any Key to continue...");
getch(); return 0;}
if(i!=k)
{ if((m=0-a1[i][k]/a1[k][k])==0) {i++; continue;}
else {for(j=k+1;j<=n;j++)
a1[i][j]=a1[i][j]+a1[k][j]*m;
b1[i]=b1[i]+b1[k]*m;}
i++;}
else i++; }}
/* yi xia ji suan x zhi */
for(i=n;i>=1;i--)
x1[i]=b1[i]/a1[i][i];
return 1;}
double computed(double **a,int h,int l, int *c1,int n) /*计算系数行列式D值*/
{ int i, j,p=1;
double sum=0.0;
if(h==n)
sum=1.0;
else {
i=++h;
c1[l]=0;
for(j=1;j<=n;j++)
if(c1[j])
if(a[i][j]==0) p++;
else {sum=sum+a[i][j]*computed(a,i,j,c1,n)*pow(-1,1+p); p++; } c1[l]=1; }
return sum; }
void ncompute(double **a,double b[],double x[],int n,int *c,double h) /*克莱姆法计算*/
{int i,j;
double t[NUM];
for(j=1;j<=n;j++)
{ for(i=1;i<=n;i++)
{t[i]=a[i][j];a[i][j]=b[i];}
x[j]=computed(a,0,0,c,n)/h;
for(i=1;i<=n;i++)
a[i][j]=t[i]; }
}
main()
{double x[NUM];
double b[NUM];
int i,j=2,n=0;
int *c;
double he;
char m,decision;
double **a;
a=(double**)malloc(NUM*sizeof(double*));
for (i=0; i<NUM; i++)
a[i]=(double*)malloc(NUM*sizeof(double));
print();
do{
clrscr();
do{
if(n>=NUM) printf("n is too large,please input again:\n");
else
printf("Please input the total number of the equations n(n<NUM): \n");
scanf(" %d",&n);
}while(n>NUM);
if(n==0) {for(i=1; i<NUM; i++) free(a[i]);
free(a); exit(1);}
input(a,b,n);
c=(int *)malloc((n+1)*sizeof(int));
memset(c,1,(n+1)*sizeof(int));
he=computed(a,0,0,c,n);
if(fabs(he)>1e-4)
{
Other: chose();
do{
m=getche();
}while(m!='a'&&m!='b'&&m!='A'&&m!='B'&&m!='c'&&m!='C'&&m!='d'&&m!='D'&&m!='e'&&m!='E');
switch(m)
{ case 'a': ;
case 'A': j=gscompute(a,b,x,n); break;
case 'b': ;
case 'B': j=gs_ydcompute(a,b,x,n); break;
case 'c': ;
case 'C': j=ddcompute(a,b,x,n); break;
case 'd': ;
case 'D': j=1; ncompute(a,b,x,n,c,he); break;
case 'e': ;
case 'E': j=2; break;
default: j=2; break;
}
if(j==1)
{ clrscr();
printf("\n\n\n\n");
printf(" D=%.4f \n",he);
for(i=1;i<=n;i++)
{if(i%5==0) printf("\n");
printf(" %.4f ",x[i]);}
}
else if(j==0)
{printf(" \nThese equtions can't be solve is this way.\nPlease chose the other way."); goto Other;}
else {for(i=1; i<NUM; i++) free(a[i]);
free(a);
free(c);
exit(1);}
}
else printf(" \n\n\tD=%.4f\n This linear equations hasn't accurate answer!",he);
printf(" \n Do you want to continue?(Y/N) \n");
do{
decision=getchar();}while(decision!='y'&&decision!='Y'&&decision!='n'&&decision!='N');
}while(decision=='y'||decision=='Y');
for(i=1; i<NUM; i++) free(a[i]);
free(a);
free(c);}
