#include <stdlib.h>
#include <malloc.h>
void MatrixOpp(double *A, int m, int n, double* invmat)
void MatrixInver(double *A, int m, int n, double* invmat)
double Surplus(double A[], int m, int n)
int matrix_inv(double* p, int num, double* invmat)
void MatrixOpp(double A[], int m, int n, double* invmat)
{
int i, j, x, y, k
double *SP = NULL, *AB = NULL, *B = NULL, X
SP = (double *) malloc(m * n * sizeof(double))
AB = (double *) malloc(m * n * sizeof(double))
B = (double *) malloc(m * n * sizeof(double))
X = Surplus(A, m, n)
X = 1 / X
for (i = 0 i < m i++)
for (j = 0 j < n j++)
{
for (k = 0 k < m * n k++)
B[k] = A[k]
{
for (x = 0 x < n x++)
B[i * n + x] = 0
for (y = 0 y < m y++)
B[m * y + j] = 0
B[i * n + j] = 1
SP[i * n + j] = Surplus(B, m, n)
AB[i * n + j] = X * SP[i * n + j]
}
}
MatrixInver(AB, m, n, invmat)
free(SP)
free(AB)
free(B)
}
void MatrixInver(double A[], int m, int n, double* invmat)
{
int i, j
double *B = invmat
for (i = 0 i < n i++)
for (j = 0 j < m j++)
B[i * m + j] = A[j * n + i]
}
double Surplus(double A[], int m, int n)
{
int i, j, k, p, r
double X, temp = 1, temp1 = 1, s = 0, s1 = 0
if (n == 2)
{
for (i = 0 i < m i++)
for (j = 0 j < n j++)
if ((i + j) % 2)
temp1 *= A[i * n + j]
else
temp *= A[i * n + j]
X = temp - temp1
}
else
{
for (k = 0 k < n k++)
{
for (i = 0, j = k i < m, j < n i++, j++)
temp *= A[i * n + j]
if (m - i)
{
for (p = m - i, r = m - 1 p > 0 p--, r--)
temp *= A[r * n + p - 1]
}
s += temp
temp = 1
}
for (k = n - 1 k >= 0 k--)
{
for (i = 0, j = k i < m, j >= 0 i++, j--)
temp1 *= A[i * n + j]
if (m - i)
{
for (p = m - 1, r = i r < m p--, r++)
temp1 *= A[r * n + p]
}
s1 += temp1
temp1 = 1
}
X = s - s1
}
return X
}
int matrix_inv(double* p, int num, double* invmat)
{
if (p == NULL || invmat == NULL)
{
return 1
}
if (num > 10)
{
return 2
}
MatrixOpp(p, num, num, invmat)
return 0
}
int main()
{
int i, j
int num
double *arr=NULL
double *result=NULL
int flag
scanf("%d",&num)
arr=(double *)malloc(sizeof(double)*num*num)
result=(double *)malloc(sizeof(double)*num*num)
printf("请输入%d*%d矩阵：\n", num, num)
for (i = 0 i < num i++)
{
for (j = 0 j < num j++)
{
scanf("%lf", &arr[i * num + j])
}
}
flag = matrix_inv(arr, num, result)
if(flag==0)
{
printf("逆矩阵为：\n")
for (i = 0 i < num * num i++)
{
printf("%lf\t ", *(result + i))
if (i % num == (num - 1))
printf("\n")
}
}
else if(flag==1)
{
printf("p/q为空\n")
}
else
{
printf("超过最大维数\n")
}
system("PAUSE")
free(arr)
free(result)
return 0
}
#include <stdio.h>#include <stdlib.h>#include <malloc.h>void MatrixOpp(double *A, int m, int n, double* invmat)void MatrixInver(double *A, int m, int n, double* invmat)double Surplus(double A[], int m, int n)int matrix_inv(double* p, int num, double* invmat)void MatrixOpp(double A[], int m, int n, double* invmat){ int i, j, x, y, k double *SP = NULL, *AB = NULL, *B = NULL, X SP = (double *) malloc(m * n * sizeof(double)) AB = (double *) malloc(m * n * sizeof(double)) B = (double *) malloc(m * n * sizeof(double)) X = Surplus(A, m, n) X = 1 / X for (i = 0i <mi++) for (j = 0j <nj++) { for (k = 0k <m * nk++) B[k] = A[k] { for (x = 0x <nx++) B[i * n + x] = 0 for (y = 0y <my++) B[m * y + j] = 0 B[i * n + j] = 1 SP[i * n + j] = Surplus(B, m, n) AB[i * n + j] = X * SP[i * n + j] } } MatrixInver(AB, m, n, invmat) free(SP) free(AB) free(B)}void MatrixInver(double A[], int m, int n, double* invmat){ int i, j double *B = invmat for (i = 0i <ni++) for (j = 0j <mj++) B[i * m + j] = A[j * n + i]}double Surplus(double A[], int m, int n){ int i, j, k, p, r double X, temp = 1, temp1 = 1, s = 0, s1 = 0 if (n == 2) { for (i = 0i <mi++) for (j = 0j <nj++) if ((i + j) % 2) temp1 *= A[i * n + j] else temp *= A[i * n + j] X = temp - temp1 } else { for (k = 0k <nk++) { for (i = 0, j = ki <m, j <ni++, j++) temp *= A[i * n + j] if (m - i) { for (p = m - i, r = m - 1p >0p--, r--) temp *= A[r * n + p - 1] } s += temp temp = 1 } for (k = n - 1k >= 0k--) { for (i = 0, j = ki <m, j >= 0i++, j--) temp1 *= A[i * n + j] if (m - i) { for (p = m - 1, r = ir <mp--, r++) temp1 *= A[r * n + p] } s1 += temp1 temp1 = 1 } X = s - s1 } return X}int matrix_inv(double* p, int num, double* invmat){ if (p == NULL || invmat == NULL) { return 1 } if (num >10) { return 2 } MatrixOpp(p, num, num, invmat) return 0}int main(){ int i, j int num double *arr=NULL double *result=NULL int flag printf("请输入矩阵维数：\n") scanf("%d",&num) arr=(double *)malloc(sizeof(double)*num*num) result=(double *)malloc(sizeof(double)*num*num) printf("请输入%d*%d矩阵：\n", num, num) for (i = 0i <numi++) { for (j = 0j <numj++) { scanf("%lf", &arr[i * num + j]) } } flag = matrix_inv(arr, num, result) if(flag==0) { printf("逆矩阵为：\n") for (i = 0i <num * numi++) { printf("%lf\t ", *(result + i)) if (i % num == (num - 1)) printf("\n") } } else if(flag==1) { printf("p/q为空\n") } else { printf("超过最大维数\n") } system("PAUSE") free(arr) free(result) return 0}//源程序如下#include<stdio.h>#include<conio.h>
#include<string.h>
#include<iostream.h>
#include<stdlib.h>
#include<math.h>
#define max 100void inputstyle(int *) //输入函数
void input(int **,int) //输入函数
long danx(int **,int)
int sgnx(int)
void martx(int **,int)int main(void)
{
int style=0,i=0
int matrix[max][max],*p[max]
for(i=0i<maxi++)*(p+i)=matrix[i] //*(p+i)是指针，指向第i个字符串
char exit1=' '
while(exit1!='E'&&exit1!='e'){ printf("求n阶矩阵的逆\n") inputstyle(&style)
input(p,style)
printf("原矩阵为:\n")
for(i=0i<stylei++){
for(int j=0j<stylej++){
printf("%4d",matrix[i][j])
}
printf("\n")
}
martx(p,style)
printf("\n")
printf("Exit=e Continue=Press any key\n")
cin>>exit1
fflush(stdin)
printf("\n\n")}
return(0)
} void input(int **p,int n){
for(int i=0i<ni++){
for(int j=0j<nj++){
printf("输入矩阵(%d行,%d列)元素:",j+1,i+1)
*(*(p+j)+i)=0
scanf("%d",*(p+j)+i)
fflush(stdin)
}
}
}void inputstyle(int *style){
do{
printf("输入矩阵n*n阶数n(0<n<%d):",max)
fflush(stdin)
scanf("%d",style)
fflush(stdin)
}while(*style<=0 &&*style>max)
}long danx(int **p,int n){
int i=0,j1=0,k1=0,j2=0,k2=0
long sum=0
int operate[max][max],*po[max]
for(i=0i<maxi++)*(po+i)=operate[i]if(n==1)return *(*(p+0)+0)
else{
for(i=0i<ni++){
for(j1=1,j2=0j1<nj1++,j2++){
k1=-1k2=-1
while(k2<n-1){
k1++
k2++
if(k1==i)k1++
*(*(po+j2)+k2)=*(*(p+j1)+k1)
}
}
/*for(int i1=0i1<n-1i1++){
for(int h1=0h1<n-1h1++){
printf("(%d,%d)%d ",i1,h1,*(*(po+h1)+i1))
}
printf("\n")
}*/
sum+=*(*(p+0)+i) * sgnx(1+i+1) * danx(po,n-1)
}
return sum
}
}int sgnx(int i){
if(i%2==0)return(1)
else return(-1)
}void martx(int **p,int n){
int i=0,j=0,j1=0,k1=0,j2=0,k2=0,num=0
int tramform[max][max]
int operate[max][max],*po[max]
for(i=0i<maxi++)*(po+i)=operate[i]
num=danx(p,n)
if(num==0)printf("矩阵不可逆\n")
else{
if(n==1)printf("矩阵的逆为: 1/%d\n",num)
else{
printf("矩阵的逆为: 系数 1/%d *\n",num)
for(i=0i<ni++){
for(j=0j<nj++){
j1=-1j2=-1
while(j2<n-1){
j1++j2++
if(j1==j)j1++ k1=-1k2=-1
while(k2<n-1){
k1++
k2++
if(k1==i)k1++
*(*(po+j2)+k2)=*(*(p+j1)+k1)
}
}
tramform[i][j]=sgnx(2+i+j) * danx(po,n-1)
}
}
for(i=0i<ni++){
for(j=0j<nj++){
printf("%4d",tramform[i][j])
}
printf("\n")
}
}
}
}
//运行结果//希望对你有帮助
