Codeforces 509E 想法

Pretty Song

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 si be the i-th character of string s, and si..j be the substring of word s, staring at the i-th character and ending at the j-th character (sisi + 1... sj, i ≤ j).

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

The prettiness of s equals

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.

Examples

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, theprettiness of the song equals to 28.

题意：

定义一个字符串的优美度为这个字符串中元音字母所占的比例。
给定一个字符串，对这个字符串所有子串的优美度求和。

题解：我们算每个字符对所有子串的贡献  很容易得出这样的结果

1+1/2+1/3+1/4
     1/2+1/3+1/4+1/5
             1/3+1/4+1/5+1/6

那么问题就变成了求解这个平行四边形的和

预处理三角形的阶层和

把下面补齐  上面补齐  变成一个直角三角形

那么就是拿大三角形的阶层和减去上面的再减去左下角的

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char s[500005];double sum[500005];int len;void init(){int i=1;double d=0;for(i=1;i<=500000;i++){d+=1.0/i;sum[i]=sum[i-1]+d;}}int judge(char t){if(t=='A'||t=='I'||t=='E'||t=='U'||t=='O'||t=='Y')return 1;return 0;}double solve(int t){return sum[len]-sum[len-t]-sum[t-1];//整个-上面-左下角}int main(){init();scanf("%s",s+1);len=strlen(s+1);int i;double ans=0;for(i=1;i<=len;i++){if(judge(s[i])){ans+=solve(i);}}printf("%.8f\n",ans);return 0;}