CF205E-Little Elephant and Furik and Rubik

CF205E-Little Elephant and Furik and Rubik

Description

Little Elephant loves Furik and Rubik, who he met in a small city Kremenchug.

The Little Elephant has two strings of equal length a and b, consisting only of uppercase English letters. The Little Elephant selects a pair of substrings of equal length — the first one from string a, the second one from string b. The choice is equiprobable among all possible pairs. Let’s denote the substring of a as x, and the substring of b — as y. The Little Elephant gives string x to Furik and string y — to Rubik.

Let’s assume that $f(x, y)$ is the number of such positions of i ($1 ≤ i ≤ |x|$), that $x_i = y_i$ (where $|x|$ is the length of lines x and y, and $x_i, y_i$ are the i-th characters of strings x and y, correspondingly). Help Furik and Rubik find the expected value of $f(x, y)$.

Input

The first line contains a single integer n ($1 ≤ n ≤ 2·10^5$) — the length of strings a and b. The second line contains string a, the third line contains string b. The strings consist of uppercase English letters only. The length of both strings equals $n$.

Output

On a single line print a real number — the answer to the problem. The answer will be considered correct if its relative or absolute error does not exceed $10^{-6}$.

题解

$E(n)=\sum P_i*f(x,y)$

$P_i$表示每种字串$x,y$出现的可能

同时这个等式也可以表示成如下的形式：

$E(n)=\sum P_i*\sum_{l}^{j=0}x_j==y_j$

$x_j==y_j$为1否则为0

$l$为子串$x$的长度

然后转化为

$E(n)=\sum\sum_{l}^{i=0}P_i*(x_j==y_j)$

因为每种字串的出现的可能都是相等的

即

$P_1==P_2==...P_n$

所以

$E(n)=P*\sum\sum_{l}^{j=0}x_j==y_j$

我们可以看出本质上题目要求的就是所有子串中$x_i==y_i$的个数

与概率无关

那么我们可以枚举字符$c$，看能构造多少个子串$x_i==c，y_i==c$

这里我们采用

比如说祖串$X:abav,Y:bacd$,对于$Y_2$也就是$a$来说，我们可以匹配$X_1$，因为$X_1$为第一$X$第一个字符，所以无法向左扩展，向右可以扩展2个长度，我们还可以匹配$X_3$,向左扩展1个长度，向右扩展1个长度，那么对于$a$来说，总共可以构造出的字串的个数为，$1*3+2*2$个。

但是我们发现这样还是没有办法解决问题，直接搜索匹配字符的复杂度为$O(N^2)$的，对于$2*10^5$的数据肯定超时，那么如何优化时间复杂度呢？

经过大仙的提醒我们可以观察到,对于$Y_i$，对于在X串相同位置之前的匹配的字符$X_j$向左扩展的长度都是有$X_j$决定的即为j，向右扩展的长度都是由$Y_i$决定的即为$i$,对于$Y_i$之后的字符我们能够找到类似的性质，那么我们就可以预处理$\sum j(Y_j==X_i,j<=i)$，和$\sum j(Y_j==X_i,j>=i)$，然后通过$O(N)$的复杂度求解。

PS:注意$X_i$不要算重复了

代码

/*********************************
Author: Toudsour
Created Time: 六  8/29 14:56:25 2015
File Name:CF205E.cpp
*********************************/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char StrA[200100];
char StrB[200100];
long long CountP[26][200100];
long long CountE[26][200100];
int main()
{
    int  N;
    scanf("%d",&N);
    scanf("%s",&StrA[1]);
    scanf("%s",&StrB[1]);
    for(int i=1;i<=N;i++)
    {
        int A=StrA[i]-'A';
        CountP[A][i]=i;
    }
    for(int i=N;i>=1;i--)
    {
        int A=StrA[i]-'A';
        CountE[A][i]=N+1-i;
    }
    //for(int i=1;i<N;i++)
    //    cout<<CountP[0][i]<<" ";
    //cout<<endl;
    for(int i=1;i<=N;i++)
        for(int j=0;j<26;j++)
            CountP[j][i]+=CountP[j][i-1];
    for(int i=N-1;i>=1;i--)
        for(int j=0;j<26;j++)
            CountE[j][i]+=CountE[j][i+1];
    double Sum=0;
    for(int i=1;i<=N;i++)
    {
        //cout<<i<<endl;
        int A=StrB[i]-'A';
        Sum+=CountP[A][i]*(N+1-i);
        //cout<<(N-i+1)<<" "<<CountP[A][i]<<endl;
        Sum+=CountE[A][i+1]*i;
        //cout<<i<<" "<<CountE[A][i]<<endl;
    }
    double Ans=Sum;
    double Temp=N;
    Ans=Ans/(Temp*(Temp+1)*(2*Temp+1)/6);
    printf("%.9f\n",Ans);
    return 0;
}