[贪心]cf226bBear and Strings

最新推荐文章于 2020-02-06 10:41:01 发布

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最新推荐文章于 2020-02-06 10:41:01 发布

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分类专栏： ACM

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本文介绍了一个算法问题：在给定字符串中计算包含子串bear的所有子串对的数量。通过记录每个bear出现的位置并利用这些位置计算符合条件的子串对数量，提供了一种高效的解决方案。

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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) = s_is_i + 1... s_j 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 s_k = b, s_k + 1 = e, s_k + 2 = a, s_k + 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.

Sample test(s)

Input

bearbtear

Output

Input

bearaabearc

Output

Note

In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).

In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).

注意bear里四个字母都不一样，就非常简单了。

#include <cstdio>
#include <cstring>

char str[5100];
int next[30];
int pos[5100];

int main()
{
	//freopen("b.in","r",stdin);
	//freopen("b.out","w",stdout);

	scanf("%s",str+1);
	int len = strlen(str+1);
	int cnt = 0;

	for (int i=1;i<=len;i++)
	{
		if (str[i]=='b'&&str[i+1]=='e'&&str[i+2]=='a'&&str[i+3]=='r')
		{
			pos[++cnt] = i;
		}
	}

	int ans = 0;
	for (int i=1;i<=cnt;i++)
	{
		ans += (pos[i]-pos[i-1])*(len-pos[i]-2);
	}
	printf("%d",ans);
	return 0;
}