Algorithm Analysis

We draw a rectangular frame of $n\times m$ as follows:

• In the first row, we output $m$ asterisks;

• We output $n-2$ rows. In each row, we output an asterisk, $m-2$ spaces, and another asterisk, but only if $m>1$;

• In the last $n$-th row, we output $m$ asterisks, but only if $n>1$.

Algorithm Implementation

Read the input values $n$ and $m$.

scanf("%d %d", &n, &m);

Output the first row – $m$ asterisks.

for (i = 0; i < m; i++)
    printf("*");
printf("\n");

Output $n-2$ rows. In each row, output an asterisk, $m-2$ spaces, and another asterisk, but only if $m>1$.

for (i = 0; i < n - 2; i++)
{
    printf("*");
    for (j = 0; j < m - 2; j++)
        printf(" ");
    if (m > 1)
        printf("*");
    printf("\n");
}

Output the last row – $m$ asterisks. Output it only if $n>1$.

if (n > 1)
{
    for (i = 0; i < m; i++)
        printf("*");
    printf("\n");
}