http://codeforces.com/problemset/problem/509/E
求一个字符串的所有字串的权值和,每个字串的权值为元音字母的个数比上字串的长度
将字串转化为01串,那么区间[l,r]的字串的权值为(s[r]-s[l-1])/(r-l+1),枚举长度k,则所有字串的权值和为
Sigma(1/k *(s[k]-[s0] + s[k+1]-s[1]+...s[n]-s[n-k]) ) 一式
令sum[i]=s[0]+s[1]+s[2]+...+s[i]
则一式转化为sum[n]-sum[k-1]-sum[n-k]
#include <cstdio>
#include <numeric>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <bitset>
#include <queue>
#include <sstream>
#include <deque>
using namespace std;
string s="IEAOUY";
string ss,sss;
double cnt1[500007],sum[500007];
double ans=0.0;
int flag1;
int main()
{
//std::ios::sync_with_stdio(false);
//std::cin.tie(0);
//freopen("in.txt","r",stdin);
memset(cnt1,0.0,sizeof(cnt1));
memset(sum,0.0,sizeof(cnt1));
cin>>ss;
int n=ss.size();
//cout<<n<<endl;
for(int i=0;i<n;i++)
{
flag1=0;
for(int j=0;j<6;j++)
{
if(ss[i]==s[j])
{
sss+='1';
flag1=1;
break;
}
}
if(!flag1) sss+='0';
}
//cout<<sss<<endl;
if(sss[0]=='1') cnt1[0]=1;
else cnt1[0]=0;
for(int i=1;i<n;i++)
{
if(sss[i]=='1') cnt1[i]=cnt1[i-1]+1;
else cnt1[i]=cnt1[i-1];
//cout<<cnt1[i]<<endl;
}
sum[0]=cnt1[0];
for(int i=1;i<n;i++)
{
sum[i]=sum[i-1]+cnt1[i];
//cout<<sum[i]<<endl;
}
for(int i=0;i<n;i++)
{
ans+=(sum[n-1]-sum[i-1]-sum[n-i-2])/((i+1)*1.0);
//cout<<(sum[n-1]-sum[i-1]-sum[n-i-2])/((i+1)*1.0)<<endl;
}
printf("%.7f",ans);
//cout<<ans<<endl;
return 0;
}