Thread: c programming problem

I am trying to do a few practice problems from C by Dissection and can't figure out this problem at all.


The problem asks "write a program that reads n integers into an array, then prints on a separate line the value of each distinct element along with the number of times it occurs. the values should be printed in descending order."

it gives the example that the user inputs:

-7 -7 3 3 5 -7

and the program prints:

5 occurs 1 times
3 occurs 2 times
-7 occurs 3 times


it also asks that I use dynamic memory allocation and pointer arithmetic


if anyone can figure out this program that would be greatly appreciated.

Isn't it the point of the exercise that you figure it out?

Give it your best try, if you get stuck post your code (in code tags, please) and maybe we can help...

If you have a copy of K&R's C, try looking at Chapter 1, Exercises 1-9 to 1-12 for examples of simple counting and reporting; also, for the core of the problem maybe consider Chapter 6.3 to 6.5 (arrays and pointers to structures), that section has a more involved set of examples which is closer to your problem, but doesn't cover the basic matter of one line per reported characteristic.

I'm sorry I cannot help you about the book that you have. K&R's C is the only C book I have and use.

Break that problem down into smaller parts. Use your flowcharting skills to solve the logic of the problem before you start typing. Try notebook paper before typing. I find it helps.

Yea, I don't have that book unfortunately.

I have the C by dissection 4th ed. and I am looking all over the book for hints or ways to do this problem and I can't find anything.

here is the code for you, friend. but i seem abit too long.

Code:

#include<stdio.h>
main()
{
	int arr[10],i,temp=0,count1=0,count2=0,count3=0,count4=0,count5=0,count6=0,count7=0,count8=0,count9=0,count10=0;
for(i=0;i<10;i++)
{
	scanf("%d",&arr[i]);
}

for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count1++;
}
for(i=0;i<10;i++)
{	temp=arr[1];
	if(arr[i]==temp)
	count2++;
}

for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count3++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count4++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count5++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count6++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count7++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count8++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count9++;
}
for(i=0;i<10;i++)
{	temp=arr[0];
	if(arr[i]==temp)
	count10++;
}


printf("%d occur %d times\n",arr[0],count1);
printf("%d occur %d times\n",arr[1],count2);
printf("%d occur %d times\n",arr[2],count3);
printf("%d occur %d times\n",arr[3],count4);
printf("%d occur %d times\n",arr[4],count5);
printf("%d occur %d times\n",arr[5],count6);
printf("%d occur %d times\n",arr[6],count7);
printf("%d occur %d times\n",arr[7],count8);
printf("%d occur %d times\n",arr[8],count9);
printf("%d occur %d times\n",arr[9],count10);
}

What do you do if n > 1000?

You might want to think about cutting that down...

If you use an array for your count and work with two nested loops, you could probably do that whole function in about 10 lines of code.

Something like this pseudocode...

Code:

array 10
count 10

for outerloop = 0 to 9
  for innerloop = 0 to 9
         test innerloop not equals outerloop
            compare  array[innerloop] to array[outerloop]
                increase count[outerloop] by 1 if equal