Simultaneous equation using gauss elimination method

Program:

#include<stdio.h>
#include<conio.h>
void main()
{
    int i, j, k, n;
    float a[20][20], x[20];
    double s, p ;
    clrscr();
    printf("Enter the number of equations : ");
    scanf("%d", &n);
    printf("\nEnter the coefficients of the equations :\n\n");
    for(i = 0 ; i < n ; i++)
    {
        for(j = 0 ; j < n ; j++)
        {
            printf("a[%d][%d] = ", i + 1, j + 1);
            scanf("%f", &a[i][j]);
        }
        printf("b[%d] = ", i + 1);
        scanf("%f", &a[i][n]);
    }
    for(k = 0 ; k < n - 1 ; k++)
    {
        for(i = k + 1 ; i < n ; i++)
        {
            p = a[i][k] / a[k][k];
            for(j = k ; j < n + 1 ; j++)
            a[i][j] = a[i][j] - p * a[k][j];
        }
    }
    x[n-1] = a[n-1][n] / a[n-1][n-1];
    for(i = n - 2 ; i >= 0 ; i--)
    {
        s = 0;
        for(j = i + 1 ; j < n ; j++)
        {
            s  = (a[i][j] * x[j]);
            x[i] = (a[i][n] - s) / a[i][i];
        }
    }
    printf("\nThe result is :\n");
    for(i = 0 ; i < n ; i++)
    printf("\nx[%d] = %.2f", i + 1, x[i]);
    getch();
}

Output:

Enter the number of equations : 3
Enter the coefficients of the equations :
a[1][1] = 10
a[1][2] = 1
a[1][3] = 1
b[1] = 12
a[2][1] = 2
a[2][2] = 10
a[2][3] = 1
b[2] = 13
a[3][1] = 1
a[3][2] = 1
a[3][3] = 5
b[3] = 7
The result is :
x[1] = 1.00
x[2] = 1.00
x[3] = 1.00