求c语言求矩阵的行列式和逆天天向哪儿
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解决时间 2021-01-28 08:41
- 提问者网友:雾里闻花香
- 2021-01-27 18:49
求c语言求矩阵的行列式和逆天天向哪儿
最佳答案
- 五星知识达人网友:你可爱的野爹
- 2021-01-27 20:16
引用型参数应该在能被定义为const的情况下,尽量定义为const 。
#define N 5
#include "stdio.h"
#include "conio.h"
int js(s,n)
int s[][N],n;
{int z,j,k,r,total=0;
int b[N][N];
if(n>2) {for(z=0;z {for(j=0;j for(k=0;k if(k>=z) b[j][k]=s[j+1][k+1];
else b[j][k]=s[j+1][k];
if(z%2==0) r=s[0][z]*js(b,n-1);
else r=(-1)*s[0][z]*js(b,n-1);
total=total+r;
}
}
else if(n==2) total=s[0][0]*s[1][1]-s[0][1]*s[1][0];
return total;
}
void n_1(s,b,n)
int s[][N],n;
float b[][N];
{int z,j,k,l,m,g,a[N][N];
for(z=0;z {l=z;
for(j=0;j { m=j;
for (k=0;k for(g=0;g { if(g>=m&&k else if(k>=l&&g else if(k>=l&&g>=m) a[k][g]=s[k+1][g+1];
else a[k][g]=s[k][g];
}
b[z][j]=js(a,n-1);
}
}
}
main()
{int a[N][N];
float b[N][N];
int r,z,j;
float temp;
//clrscr();
printf("Input original data:\n");
for(z=0;z for(j=0;j scanf("%d",&a[z][j]);
printf("\nPress Enter continue......");
getchar();
//gotoxy(1,1);
printf("The original matrix is:\n");
for(z=0;z {for(j=0;j printf("%5d",a[z][j]);
printf("\n");
}
r=js(a,N);
printf("\nThe original matrix hanglieshi is:|A|==%d\n",r);
if (r==0) printf("Because |A|==0,the original matrix have no nijuzhen!");
else
{n_1(a,b,N);
for(z=0;z for(j=0;j if((z+j)%2!=0 && b[z][j]!=0) b[z][j]=-b[z][j];
for(z=0;z for(j=z+2;j {temp=b[z][j];
b[z][j]=b[j][z];
b[j][z]=temp;
}
printf("Because |A|!=0,the original matrix have nijuzhen!\n");
printf("The bansuijuzhen A* is:\n");
for(z=0;z {for(j=0;j printf("%4.0f\t",b[z][j]);
printf("\n");
}
for(z=0;z for(j=0;j b[z][j]=b[z][j]/r;
printf("\nThe nijuzhen is:(A*)/|A|(|A|=%d)\n",r);
for(z=0;z {for(j=0;j printf("%8.3f",b[z][j]);
printf("\n");
}
}
}
#define N 5
#include "stdio.h"
#include "conio.h"
int js(s,n)
int s[][N],n;
{int z,j,k,r,total=0;
int b[N][N];
if(n>2) {for(z=0;z
else b[j][k]=s[j+1][k];
if(z%2==0) r=s[0][z]*js(b,n-1);
else r=(-1)*s[0][z]*js(b,n-1);
total=total+r;
}
}
else if(n==2) total=s[0][0]*s[1][1]-s[0][1]*s[1][0];
return total;
}
void n_1(s,b,n)
int s[][N],n;
float b[][N];
{int z,j,k,l,m,g,a[N][N];
for(z=0;z
for(j=0;j
for (k=0;k
else a[k][g]=s[k][g];
}
b[z][j]=js(a,n-1);
}
}
}
main()
{int a[N][N];
float b[N][N];
int r,z,j;
float temp;
//clrscr();
printf("Input original data:\n");
for(z=0;z
printf("\nPress Enter continue......");
getchar();
//gotoxy(1,1);
printf("The original matrix is:\n");
for(z=0;z
printf("\n");
}
r=js(a,N);
printf("\nThe original matrix hanglieshi is:|A|==%d\n",r);
if (r==0) printf("Because |A|==0,the original matrix have no nijuzhen!");
else
{n_1(a,b,N);
for(z=0;z
for(z=0;z
b[z][j]=b[j][z];
b[j][z]=temp;
}
printf("Because |A|!=0,the original matrix have nijuzhen!\n");
printf("The bansuijuzhen A* is:\n");
for(z=0;z
printf("\n");
}
for(z=0;z
printf("\nThe nijuzhen is:(A*)/|A|(|A|=%d)\n",r);
for(z=0;z
printf("\n");
}
}
}
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