#print
Write a program to read a list of positive numbers
and sort them into ascending order.  Print
the sorted list of numbers one per line
as five digit numbers (%5d in printf).
Stop reading numbers when getnum returns -1.
Compile and test your program; then type "ready".
#once #create Ref
    1
    3
    4
    9
   11
   12
   13
   14
   15
   16
   17
   20
   34
   71
  200
  225
  250
  275
  300
 4095
 7111
16384
#once cp %s/getnum.o .
#once #create input
4 20 3 200 16384 4095 71 11 12 13 14
15 16 17 34 9 7111 300 275 250 225 1
#user
a.out  <input >xxx
#cmp xxx Ref
#succeed
main()
{
	int n;
	int list[1000];

	n = getlist(list);
	shellsort(list, n);
	printlist(list, n);
}

getlist(list)
int list[];
{
	int n;

	n = 0;
	while ((list[n]=getnum()) >= 0)
		n++;
	return(n);
}

/* this is a shell sort, stripped down to process a list
   of integers only.  Although you probably don't know
   how to write this offhand, you should know where to find
   it - it is only marginally more code than a bubble sort
   and much faster (n**1.5 vs. n**2) in time. */
shellsort(v, n)  /* sort v[0]...v[n-1] into increasing order */
int v[], n;
{
    int gap, i, j, temp;

    for (gap = n/2; gap > 0; gap /= 2)
        for (i = gap; i < n; i++)
            for (j=i-gap; j>=0 && v[j]>v[j+gap]; j-=gap) {
                temp = v[j];
                v[j] = v[j+gap];
                v[j+gap] = temp;
            }
}

printlist(list, n)
int list[], n;
{
	int i;
	for(i=0; i<n; i++)
		printf("%5d\n",list[i]);
}
/* this is a crummy bubble sort which
   would work perfectly well for this
   problem but can not be recommended
   for large jobs. 
sortlist()
{
	int i, j, k;

	for(i=0; i<n; i++)
		for(j=n-1; j>0; j--)
			if (list[j-1] > list[j]) {
				k = list[j];
				list[j] = list[j-1];
				list[j-1] = k;
			}
}
 ****/
#log
#next
30.1a 10