Tinkoff Internship Warmup Round 2018 and Codeforces Round #475 (Div. 2)C. Alternati(费马小定理求逆元+等比数列求和)

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:

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

.

题解：
n很大，但是K只有1e5,因此可以算好两头，中间的
每长度为k的一段可以看成一个数，每个数之间成
等比数列，因此可以直接用等比数列求和公式。
代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=100005;
const ll mod=1e9+9;
char t[maxn];
ll quick(ll a,ll b)
{
    ll r=1;
    while(b)
    {
        if(b%2)r=r*a%mod;
        a=a*a%mod;
        b=b/2;
    }
    return r%mod;
}
int main()
{
    ll n,a,b,k;
    scanf("%lld%lld%lld%lld",&n,&a,&b,&k);
    scanf("%s",&t);
    ll sum=0;
    if(n<k)
    {
        for(ll i=0; i<=n; i++)
        {
            ll s=quick(a,n-i)*quick(b,i)%mod;
            if(t[i]=='+')
                sum=(sum+s)%mod;
            else
                sum=(sum-s+mod)%mod;
            //printf("%lld\n",s);
        }
    }
    else
    {
        ll i;
        for(i=0; i<k; i++)
        {
            ll s=quick(a,n-i)*quick(b,i)%mod;
            if(t[i]=='+')
                sum=(sum+s)%mod;
            else
                sum=(sum-s+mod)%mod;
        }
        ll s=sum;
        i--;
        /*for( ;i+k<=n;i+=k)
        {
            s=(s*quick(quick(a,k),mod-2)%mod)*quick(b,k)%mod;
            sum=(sum+s+mod)%mod;
        }*/
        ll q=(quick(quick(a,k),mod-2)%mod)*quick(b,k)%mod;
        ll m=(n-i)/k;
        if(q==1)
            sum=(sum+m*s)%mod;
        else
            sum=(((s*quick(q,m+1)-s)%mod*quick(q-1,mod-2))%mod)%mod;
        int j=0;
        i+=m*k;
        for(i++;i<=n;i++,j++)
        {
            if(t[j]=='+')
            {
                sum=(sum+quick(n-i,i))%mod;
            }
            else
                sum=(sum-quick(n-i,i)+mod)%mod;
        }
    }
    printf("%lld\n",sum);
    return 0;
}