我讨厌线性代数

作弊器送给志同道合的孩纸
用最笨的方法写的

只能解4阶的行列式

#include<stdio.h>
int main()
{
int a[4][4];
int k,j;
int x,y,z;
k=0;j=0;
for(;j<4;)
{
for(;k<4;)
{
printf("please input (%d,%d) number\n",k+1,j+1);
scanf("%d",&a[j][k]);
k+=1;
}
k=0;
j+=1;
}x=(a[0][0])*(a[1][1])*(a[2][2])*(a[3][3])+
(a[0][0])*(a[1][2])*(a[2][3])*(a[3][1])+
(a[0][0])*(a[1][3])*(a[2][1])*(a[3][2])+
(a[0][1])*(a[1][0])*(a[2][3])*(a[3][2])+
(a[0][1])*(a[1][2])*(a[2][0])*(a[3][3])+
(a[0][1])*(a[1][3])*(a[2][2])*(a[3][0])+
(a[0][2])*(a[1][0])*(a[2][1])*(a[3][3])+
(a[0][2])*(a[1][3])*(a[2][0])*(a[3][1])+
(a[0][2])*(a[1][1])*(a[2][3])*(a[3][0])+
(a[0][3])*(a[1][0])*(a[2][2])*(a[3][1])+
(a[0][3])*(a[1][1])*(a[2][0])*(a[3][2])+
(a[0][3])*(a[1][2])*(a[2][1])*(a[3][0]);
y=(a[0][0])*(a[1][1])*(a[2][3])*(a[3][2])+
(a[0][0])*(a[1][2])*(a[2][1])*(a[3][3])+
(a[0][0])*(a[1][3])*(a[2][2])*(a[3][1])+
(a[0][1])*(a[1][0])*(a[2][2])*(a[3][3])+
(a[0][1])*(a[1][2])*(a[2][3])*(a[3][0])+
(a[0][1])*(a[1][3])*(a[2][0])*(a[3][2])+
(a[0][2])*(a[1][0])*(a[2][3])*(a[3][1])+
(a[0][2])*(a[1][1])*(a[2][0])*(a[3][3])+
(a[0][2])*(a[1][3])*(a[2][1])*(a[3][0])+
(a[0][3])*(a[1][0])*(a[2][1])*(a[3][2])+
(a[0][3])*(a[1][1])*(a[2][2])*(a[3][0])+
(a[0][3])*(a[1][2])*(a[2][0])*(a[3][1]);
z=x-y;
printf("+%d-%d\n",x,y);
printf("the answer is %d",z);
return(0);
}