codeforces Gym 100187B B. A Lot of Joy

B. A Lot of Joy

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100187/problem/A

Description

Two boys Gena and Petya wrote on two strips of paper the same string s that consisted of lowercase Latin letters. Then each boy took one strip, cut it into small pieces so that each piece contained exactly one letter and put all pieces into his pocket. After that Gena and Petya repeated the following procedure until their pockets became empty: they took one piece from their pockets and rejoiced if the letters on these pieces were equal.

Determine the expected number of times the boys rejoiced during this process.

Input

The input contains the only string s which was initially written on Gena's and Petya's strips. The string has length between 1 and 200000, inclusively, and consists of lowercase Latin letters.

Output

Output the only real number — the expected number of times Gena and Petya rejoiced during their business. The absolute or relative error of the answer should not exceed 10 - 9.

Sample Input

abc

Sample Output

1.000000000000000

HINT

题意

每个人都有一组相同的字符串，如果同时拿出一样的字符串的话，愉悦值+1，问你愉悦值的期望是多少

题解：

我也不知道为什么答案是这个，我只是随便猜的……

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200101
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

double dp;
int a[];
int b[];
int main()
{
    string s;
    cin>>s;
    for(int i=;i<s.size();i++)
    {
        a[s[i]-'a']++;
    }
    for(int i=;i<s.size();i++)
    {
        dp+=a[s[i]-'a']*1.0/s.size()*1.0;
    }
    printf("%.15f",dp);
}