Codeforces Round #475 (Div. 2) C. Alternating Sum[逆元乘法 取代 分数计算]

C. Alternating Sum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0,s_1,\dots ,s_n$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k\le i\le n$ it's satisfied that $s_i=s_{i-k}$.

Find out the non-negative remainder of division of $\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i$ by ${10}^9+9$.

Note that the modulo is unusual!

Input

The first line contains four integers $n,a,b$ and $k$ $(1\le n\le {10}^9,1\le a,b\le {10}^9,1\le k\le {10}^5)$.

The second line contains a sequence of length $k$ consisting of characters '+' and '-'.

If the $i$-th character (0-indexed) is '+', then $s_i=1$, otherwise $s_i=-1$.

Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.

Output

Output a single integer — value of given expression modulo ${10}^9+9$.

Examples

input

Copy

2 2 3 3
+-+

output

Copy

7

input

Copy

4 1 5 1
-

output

Copy

999999228

Note

In the first example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)$ = $2^2{3}^0-2^1{3}^1+2^0{3}^2$ = 7

In the second example:

$\left(\sum _{i=0}^n{s}_i{a}^{n-i}{b}^i\right)=-1^4{5}^0-1^3{5}^1-1^2{5}^2-1^1{5}^3-1^0{5}^4=-781\equiv 999999228(\mathrm{mod}{10}^9+9)$.

题意：告诉你前k个的符号+1,-1 。 对于 i>=k的符号=i-k;

思路：

1.对于前k个符号，每个符号对应后面的数列都是一个等比数列 q > 1。

2.for循环一遍前k个符号，等比公式。 

3.问题在于，解决这个等比数列如果不取模就涉及浮点数。因此正确做法使用逆元，逆元在数论中很重要，解决了在取模情况下的分数问题。

有逆元学习需要的，有学长的博客推荐 逆元学习 写得非常好。

用快速幂求逆元时 inva=qm(a,MOD-2)，要求a与MOD互质，其中题目给的mod=1e9+9是个质数

归结： 核心思想， 根据取模→用逆元乘法 取代 分数运算

以下是AC代码，复杂度O(log(max(a,b))*k)

还有一种计算方法，连续k个为一个集合，和后面是等比数列

#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;

const int N=1e5+5;
const int MOD=1e9+9;
const int INF=0x3f3f3f3f;

ll qm(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1) ans=(ans*a%MOD);
        a=(a*a)%MOD;
        b>>=1;
    }
    return ans%MOD;
}
char s[N];
int main(void){
    ll n,a,b,k;
    ll ans=0;
    cin >> n>>a>>b>>k;
    ll head=qm(a,n);
//    cout <<"head="<<head<<endl;
    ll inv_a=qm(a,MOD-2);
    ll gap=qm(b,k)*qm(inv_a,k)%MOD;
    string s;cin>>s;
//    cout <<"gap="<<gap<<endl;
    for(int i=0;i<(int)s.size();++i){
        int flag=1;
        if(s[i]=='-')  flag=-1;
        if(gap==1)  ans+=flag*(n+1)/k%MOD *head,ans= ((ans%MOD)+MOD)%MOD;
        else{
            ans+=flag*head*qm(1-gap,MOD-2)%MOD*(1-qm(gap,(n+1)/k))%MOD;
            ans= ((ans%MOD)+MOD)%MOD;
        }
        head=head*b%MOD*inv_a%MOD;
//        cout <<ans <<endl;
    }
    cout << ans << endl;
    return 0;
}