Codeforces Round #289 (Div. 二, ACM ICPC Rules)E. Pretty Song

Codeforces Round #289 (Div. 2, ACM ICPC Rules)E. Pretty Song

When Sasha was studying in the seventh grade, he started listening to music a lot. In order to evaluate which songs he likes more, he introduced the notion of the song's prettiness. The title of the song is a word consisting of uppercase Latin letters. The prettiness of the song is the prettiness of its title.

Let's define the simple prettiness of a word as the ratio of the number of vowels in the word to the number of all letters in the word.

Let's define the prettiness of a word as the sum of simple prettiness of all the substrings of the word.

More formally, let's define the function vowel(c) which is equal to 1, if c is a vowel, and to 0 otherwise. Let s_i be the i-th character of string s, and s_i..j be the substring of word s, staring at the i-th character and ending at the j-th character (s_is_i + 1... s_j, i ≤ j).

Then the simple prettiness of s is defined by the formula:

$Codeforces Round #289 (Div. 二, ACM ICPC Rules)E. Pretty Song$

The prettiness of s equals

$Codeforces Round #289 (Div. 二, ACM ICPC Rules)E. Pretty Song$

Find the prettiness of the given song title.

We assume that the vowels are I, E, A, O, U, Y.

Input

The input contains a single string s (1 ≤ |s| ≤ 5·10⁵) — the title of the song.

Output

Print the prettiness of the song with the absolute or relative error of at most 10^- 6.

Sample test(s)

Input

IEAIAIO

Output

28.0000000

Input

BYOB

Output

5.8333333

Input

YISVOWEL

Output

17.0500000

Note

In the first sample all letters are vowels. The simple prettiness of each substring is 1. The word of length 7 has 28 substrings. So, the prettiness of the song equals to 28.

我们单独考虑每一个元音字符,对于一个元音字符,先考虑以它为结尾字符的串,设这个字符的位置为i,字符串长度为len

则有 (1 / 2) + ...+ (1 / (i + 1))

再考虑以它开始的串

有 (1 / 2) + ...+ (1 / (len - i ))

单个字符的串 1个

这个字符在串中间出现:

(1 / 3) + (1 / 4) + ... + (1 / (len - i + 1)) = sum[len - i + 1] - sum[2]

(1 / 4) + (1 / 5) + ... + (1 / (len - i + 2)) = sum[len - i + 2] - sum[3]

...

(1 / (i + 2)) + (1 / (i + 3)) + ... + (1 / len) = sum[len] - sum[i + 1]

各个式子相加: sum[len - i + 1] + ... + sum[len] - (sum[2] + sum[3] + .. + sum[i + 1]) = t[len] - t[len - i - 1] - (t[i + 1] - t[1])

于是只要维护前缀和和它的前缀和就行了

/*************************************************************************
    > File Name: cf289E.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年01月31日 星期六 21时17分58秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
vector <int> p;
const int N = 5 * 101000;
char str[N];
bool flag[26];
double sum[N];
double t[N];

int main ()
{
	memset (flag, 0, sizeof(flag));
	flag['A' - 'A'] = 1;
	flag['E' - 'A'] = 1;
	flag['I' - 'A'] = 1;
	flag['O' - 'A'] = 1;
	flag['U' - 'A'] = 1;
	flag['Y' - 'A'] = 1;
	sum[1] = 1;
	for (int i = 2; i <= N - 10; ++i)
	{
		sum[i] = (1.0 / i) + sum[i - 1];
	}
	t[1] = 1;
	for (int i = 2; i <= N - 10; ++i)
	{
		t[i] = t[i - 1] + sum[i];
	}
	while (~scanf("%s", str))
	{
		int len = strlen (str);
		double ans = 0;
		for (int i = 0; i < len; ++i)
		{
			if (flag[str[i] - 'A'])
			{
				ans += sum[i + 1] - 1 + sum[len - i];
				ans += t[len] - t[len - i] - t[i + 1] + t[1];
			}
		}
		printf("%.7f\n", ans);
	}
	return 0;
}

Codeforces Round #289 (Div. 二, ACM ICPC Rules)E. Pretty Song

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