Tinkoff Internship Warmup Round 2018 and Codeforces Round #475 (Div. 2) C.Alternating Sum 题意+解析+ 答案

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:

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

这题还没过，大佬的思路（还未验证）

以下公式提出一个a^n后就有这个等比数列

b/a可以用乘法逆元c=b*a^(-1)来表示;

则有（c==1单独讨论）；

对于以k为周期的数列，

可以考虑求出每一个i+k，i+2k......i+m*k(i+m*k<=n);

则变形为X=；

令C=，则有X=

则每一个k的表达式为(sign)

快速幂加逆元，可以在O（logn）内求出每一个i；

总时间复杂度O(k*logn)

大佬代码1（已过）：

#include
using namespace std;
#define ll long long
const ll p=1e9+9;
ll Inv,C;
ll pow1(ll a,ll n)
{
    ll ans=1;
    a=a%p;
    while(n)
    {
        if(n&1)ans=ans*a%p;
        a=a*a%p;
        n>>=1;
    }
    return ans;
}
ll f(ll n)
{
    if(C==1)return n+1;
    return (pow1(C,n+1)-1+p)%p*Inv%p;
}
int main()
{
    ll n,A,B,k,c,x;
    cin>>n>>A>>B>>k;
   // if(A!=B)
        c=B*pow1(A,p-2)%p;
    C=pow1(c,k);
    if(C!=1)
        Inv=pow1(C-1,p-2);
    ll s=pow1(A,n);
    ll ans=0;
    string ss;
    cin>>ss;
    for(int i=0; i

大佬代码2（已过）：

#include
using namespace std;
#define ll long long
const ll p=1e9+9;
ll Inv,C;
ll pow1(ll a,ll n)
{
    ll ans=1;
    a=a%p;
    while(n)
    {
        if(n&1)ans=ans*a%p;
        a=a*a%p;
        n>>=1;
    }
    return ans;
}
ll f(ll n)
{
    if(C==1)return n+1;
    return (pow1(C,n+1)-1+p)%p*Inv%p;
}
int main()
{
    ll n,A,B,k,c,x;
    cin>>n>>A>>B>>k;
   // if(A!=B)
        c=B*pow1(A,p-2)%p;
    C=pow1(c,k);
    if(C!=1)
        Inv=pow1(C-1,p-2);
    ll s=pow1(A,n);
    ll ans=0;
    string ss;
    cin>>ss;
    for(int i=0; i

这题时间超时了，思路有点不对，能处理一些数据，但1e9无法很好处理

超时代码：

#include

using namespace std;
string str;
#define ll long long
ll mi(ll a,ll b)
{
    ll ans=1;
    a%=1000000009;
    while (b)
    {
        if (b%2==1) ans=ans*a%1000000009;
        b>>=1;
        a=a*a%1000000009;
    }
    return ans;
}
int main()
{
    ll n,a,b,c,t,ans=0,sum=0;
    cin >> n >> a>> b >> c ;
    cin >> str;
    for(int i=0; i<=n; i++)
    {

        if(str[i%c]=='-')
        {
            ans-=mi(a,n-i)*(mi(b,i));
            //cout << 32<< endl;
        }
        else if(str[i%c]=='+')
        {
            ans+=mi(a,n-i)*(mi(b,i));
            //cout << 33<< endl;
        }
        ans=ans%1000000009;
    }
    //cout << 3 << endl;
    if(ans<0)
    {
        ans=ans+1000000009;
    }
    cout << ans << endl;
    return 0;
}