2020 CCPC Wannafly Winter Camp Day2 A 托米的字符串（期望）

题意：

给定一个只含小写字母的字符串 $s$。求随机取一个子串元音和半元音字母 $\left(a,e,i,o,u,y\right)$ 占比的期望。

记 $a_i$表示 $s$ 的前 $i$ 个字母中元音和半元音字母个数。

记 $f_i$ 表示所有长度为 $i$ 的子串中元音和半元音字母出现次数之和。

显然有 $f_i=f_{i-1}+a_{n-i+1}-a_{i-1}$，边界条件 $f_1=a_n$

所以 $f_i=\sum_{j=1}^{n}{(a_j-a_{j-i})}=\sum_{j=n-i+1}^{n}{a_j}-\sum_{j=0}^{i-1}{a_j}$

无论是递推还是求 $a$ 的前缀和后直接算，都能在线性时间内求 $f$，

所以 $ans=\frac{2}{n(n+1)}*\sum_{i=1}^{n}{\frac{f_i}{i}}$

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll g = exgcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - a / b * y;
    return g;
}
char s[1000010];
char c[10] = " aeiouy";
ll a[1000010];
double ans;

int main()
{
    mem(a, 0);
    ss(s + 1);
    int len = strlen(s + 1);
    rep(i, 1, len)
    {
        rep(j, 1, 6)
        {
            if (s[i] == c[j])
            {
                a[i] = 1;
                break;
            }
        }
    }
    rep(i, 0, 1)
    {
        rep(j, 1, len)
        {
            a[j] += a[j - 1];
        }
    }
    rep(i, 1, len)
    {
        ans += (a[len] - a[len - i] - a[i - 1]) * 1.0 / i;
    }
    printf("%f\n", 2 * ans / len / (len + 1));
    return 0;
}