program to find the count of each word in a para

i am trying to find count of each word in a para but i am not able to get it done...

so can any one tell me how to do that..

Example Input----

Hi stack over flow is a good forum.There will be many experts in stack overflow .

ouput--

Hi---1
stack-2
overflow-2
is---1
a---1
good---1

...
...

in this way i want to get output.

This is my code...but it is not complete...after that i got struck to proceed

#include<stdio.h>
#include<conio.h>
#include<string.h>

#define NULL 0

struct wordcount
{
    char *s;
    int count;
    struct wordcount next;
}

struct checkletter
{
    char alph;
    struct wordcount next;
}

main()
{
    char *c;
    int hash[26],len,i,k=0,intm[100];
    struct checkletter complete[26];
    for(r=0;r<25;r++)
    {
        complete[r].alph=r+97;
        complete[r].next=NULL;
    }
    printf("Enter the para :");
    gets(s);
    len=strlen(s);
    for(i=0;i<len;i++)
    {       
        k=0;
        if(c[i]==' ')
        {
            for(j=i;j>m;j--)
            {
                intm[k]=c[i];
                s1=intm;
                k++;
            }
            m=k;
            hastlet=s1[0];

          开发者_开发百科  for(t=0;t<26;t++)
            {
                if(complete[t].alph==hastlet)
                {
                    if(complete[t].next==NULL)
                        complete[t].next=
                }

---

Here's a sketch:

1. Parse the paragraph into a token of words
2. Maintain a map from hash codes to counters, being careful that collisions are possible
3. For each word in the paragraph, hash the word to an integer
4. Again, being careful that collisions are possible, increment the corresponding count
5. Run through the hash table spitting out the words and the counts

Beyond that, if you have a specific question about your own implementation of the above, let us know and we can try to help. We are not going to write your code for you, though.