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analysis:

This question is immortal, we give low-equipped version of the solution and high version 2333 solution

 

Low version:

 

First, we know that such a formula. . . (To prove high version)

When a string S wherein S [1, i] = S [n - i + 1, n] when the called S [1, i] as a border of S

Ans [n] = sigma (S [1, i] is the S border) m ^ i

Ok. . .

Once you have this, we can kmp or hash solved

However, hash to get the answer can only handle S, and the answer can be done kmp handle all S prefix

Here with kmp (hash believe very simple (fog))

So we set f [i] is the i-th bit of recursion to answer

We turn first to the array fail

Then the border of the set S [1, i] is the S [1, i] itself plus S [1, fail [i]] is set border

So come recursive

f [ i ] = f [ fail [ i ] ] + m ^ i

N is the answer to the count

 

Good offices version:

 

First of all we need to know the expression builder function:

F (i) = sigma (i = 0 ... + ∞) Ai * x ^ i

We define the probability generating function generating function from which the probability of occurrence of the cost for the event i is Ai:

Then we can know:

F(1) = 1

When x is equivalent to taking 1, F (1) represents the probability and in all cases with a value of 1 is obviously correct, is the event itself

Then we consider the expected cost of E (x)

E (x) = sigma (i = 0 ... + ∞) Ai * i

Because when i do contribution is zero, so i can start from 1

Then there is an interesting thing, we have F (x) the derivative

F (x) = sigma (i = 1 ... + ∞) Ai * i * x ^ (i - 1)

We then take the F '(1) = sigma (i = 1 ... + ∞) Ai * i

And E (x) A comparison, we were surprised to find that:

E(x) = F'(1)

Is this a coincidence? Or some kind of inevitable?

This also explains the probability and expectations are necessarily linked

 

So we get to the point

Provided the function F (x) denotes the probability of the end of the length of the i, G (x) represents the length to the probability that i is not over

Then we get this equation:

F(x) + G(x) = 1 + G(x) * x

Into every consideration, in fact, the whole recursive process, you are i - the probability of a no end, is that you end position and i did not end in 'i' and the probability

This formula referred to as type 1

Then we consider solving. . .

Force plus (1 / mx) we go G (x) ^ n will be the end of the string, the string may be ended prematurely, however, because there might string before the border S

We enumerate the length of the border

First define ai = 0 or 1 represents S [1, i] whether the border S

So you can get this formula. . .

G(x) * ( 1/m * x) ^ n = sigma(i=1...n) ai * F(x) * (1/m * x) ^ ( n - i )

Hard to say this formula, but the sense of (hu) of (luan) analyze very correct 2333

Only Darknesses simple-minded, I do not know about

This formula referred to as type 2

Then we push large ♂ formula

First, a derivative of formula

F'(x) + G'(x) = G(x) + G'(x) * x

Requirements F '(1) Oh. . .

Direct access to it 2333

Then surprised to find

F'(1)=G(1)

Then take a second try

G (1) * (1 / m) * n = sigma (i = 1 ... n) and F * (1) * (1 / m) * (n - i)

Since F (1) = 1, then we have (1 / m) ^ n last addition

G (1) = sigma (i = 1 ... n) and i * m *

Therefore Ans [n] = E (x) = F '(1) = G (1) = sigma (S [1, i] to the border S) m ^ i

Get a permit. . .

 

Ha ha ha. . .

Generating function really immortal. . .

Mathematics too dishes 2333

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<queue>
#include<iostream>
 
#define maxn 1000005
#define MOD 1000000007
 
using namespace std;
 
inline int getint()
{
    int num=0,flag=1;char c;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
    return num*flag;
}
 
int n,m;
long long a[maxn],pw[maxn];
long long ans;
long long fail[maxn],f[maxn];
 
int main()
{
    m=getint(),n=getint(),f[0]=1;
    for(int i=1;i<=n;i++)a[i]=getint(),f[i]=f[i-1]*m%MOD;
    fail[0]=-1;
    for(int i=2;i<=n;i++)
    {
        for(int j=fail[i-1];j>=0;j=fail[j])
            if(a[i]==a[j+1]){fail[i]=j+1,(f[i]+=f[j+1])%=MOD;break;}
    }
    for(int i=1;i<=n;i++)printf("%lld\n",f[i]);
}

View Code

AC code can not be Oh, it outputs a direct answer all prefixes type S 2333

Code short, it proved really long 2333