URAL 1881 Long problem statement

1881. Long problem statement

Time limit: 0.5 second
Memory limit: 64 MB

While Fedya was writing the statement of the problem GOV Chronicles,
he realized that there might be not enough paper to print the statements.
He also discovered that his text editor didn't have the feature of
calculating the number of pages in a text. Then Fedya decided to write a
program that would calculate the number of pages for any given text.

Fedya knew that there were h lines on each page and w symbols in each
line. Any two neighboring words in a line were separated by exactly one
space. If there was no place for a word in a line, Fedya didn't hyphen it
but put the whole word at the beginning of the next line.

Input

The first line contains the integers h, w, and n, which are the
number of lines on a page, the number of symbols in a line, and the number
of words in the problem statement, respectively (1 ≤ h, w ≤ 100; 1 ≤ n ≤ 10 000). The statement written by Fedya is
given in the following n lines, one word per line. The words are
nonempty and consist of uppercase and lowercase English letters and
punctuation marks (period, comma, exclamation mark, and question mark);
the length of each word is at most w. The total length of all the words
is at most 10 000.

Output

Output the number of pages in the problem statement.

Sample

input	output
3 5 6 To be or not to be	2

 #include<iostream>
 #include<string.h>
 #include<stdio.h>
 #include<ctype.h>
 #include<algorithm>
 #include<stack>
 #include<queue>
 #include<set>
 #include<math.h>
 #include<vector>
 #include<map>
 #include<deque>
 #include<list>
 using namespace std;
 int main()
 {
     char s[];
     int a,b,c,d,x=,e,f=;
     scanf("%d%d%d%",&a,&b,&c);
     gets(s);
     f=strlen(s);
     e=;
     c--;
     while(c--)
     {
         gets(s);
         d=strlen(s);
         if(f+d+>b)
         {
             e++;
             f=d;
         }
         else
             f+=(d+);
         if(e>a)
         {
             e=;
             x++;
         }
     }
     printf("%d\n",x);
     return ;
 }