C语言求矩阵的逆

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <windows.h>

#define N 10

int getA(int arcs[N][N], int n)//按第一行展开计算|A|
{
    if (n == )
    {
        return arcs[][];
    }
    int ans = ;
    int temp[N][N];
    int i, j, k;
    for (i = ; i<n; i++)
    {
        for (j = ; j<n - ; j++)
        {
            for (k = ; k<n - ; k++)
            {
                temp[j][k] = arcs[j + ][(k >= i) ? k +  : k];

            }
        }
        int t = getA(temp, n - );
        if (i %  == )
        {
            ans += arcs[][i] * t;
        }
        else
        {
            ans -= arcs[][i] * t;
        }
    }
    return ans;
}
void getAStart(int arcs[N][N], int n, int ans[N][N])//计算每一行每一列的每个元素所对应的余子式，组成A*
{
    if (n == )
    {
        ans[][] = ;
        return;
    }
    int i, j, k, t;
    int temp[N][N];
    for (i = ; i<n; i++)
    {
        for (j = ; j<n; j++)
        {
            for (k = ; k<n - ; k++)
            {
                for (t = ; t<n - ; t++)
                {
                    temp[k][t] = arcs[k >= i ? k +  : k][t >= j ? t +  : t];
                }
            }

            ans[i][j] = getA(temp, n - );
            if ((i + j) %  == )
            {
                ans[i][j] = -ans[i][j];
            }
        }
    }
}

int main()
{

    int arcs[N][N];
    int astar[N][N];
    int i, j;
    int n;
    while (scanf_s("%d", &n) != EOF && n)
    {
        for (i = ; i<n; i++)
        {
            for (j = ; j<n; j++)
            {
                scanf_s("%d", &arcs[i][j]);
            }
        }
        int a = getA(arcs, n);
        if (a == )
        {
            printf("can not transform!\n");
        }
        else
        {
            getAStart(arcs, n, astar);
            for (i = ; i<n; i++)
            {
                for (j = ; j<n; j++)
                {
                    printf("%.3lf ", (double)astar[i][j] / a);
                }
                printf("\n");
            }
        }
        printf("\n");

    }
    system("pause");
    return ;
}