Codeforces Round #226 (Div. 2), problem: (B) Bear and Strings solution

Codeforces Round #226 (Div. 2), problem: (B) Bear and Strings: http://codeforces.com/contest/385/problem/B

Codeforces Round #226 (Div. 2), problem: (B) Bear and Strings editorial: http://codeforces.com/blog/entry/10514

Codeforces Round #226 (Div. 2), problem: (B) Bear and Strings solution: http://ideone.com/0uBAjE

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main() {
    int cnt, prev;
    char s[5005];
    cnt=0;
    prev=-1;
    scanf("%s", s);
    if(strlen(s)>3) for(int i=0; i<strlen(s)-3; i++) if(s[i]=='b' && s[i+1]=='e' && s[i+2]=='a' && s[i+3]=='r') {
            cnt+=(strlen(s)-i-3)*(i-prev);
            prev=i;
        }
    printf("%d", cnt);
    return 0;
}

note: you may have already realized that, subtracting "bear" substrings'  from total length of string and multiplying it with "i" index subtracted by previous "i" index will give the result.