Codeforces 385B. Bear and Strings

The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.

String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.

Help the bear cope with the given problem.

Input

The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.

Output

Print a single number — the answer to the problem.

暴力加上少量的优化

考虑s(i,j),如果s(i,j)包含bear,易得s(i,r)(r>=j)肯定都包含bear，所以直接加上后面的数量

如果s(i,j)不包含bear，那么s(i+1,j)如果有bear，肯定就是最后四个字符是bear，所以实际我们搜索只有去看s(j-3,j)这一部分就OK

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

// const int SIZE=1e6+7;


char s[5005];
const char* tar = "bear";
int total = 0;
int main(){
  // freopen("test.in","r",stdin);
  scanf("%s",s);
  int len = strlen(s);
  for (int i=0;i<len;i++){
    for (int j=i+3;j<len;j++){
      char temp = s[j+1];
      s[j+1] = '';
      if (strstr(s+j-3,tar) != NULL){
        total += len - j;
        s[j+1] = temp;
        break;
      }
      s[j+1] = temp;
    }
  }
  printf("%d
",total);
  return 0;
}