Codeforces Round #475 (Div. 2) C. Alternating Sum

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本文介绍了一种解决特定数学问题的算法，该问题要求计算一个周期性正负交替序列的等比数列求和，并对其结果进行取模运算。文章详细阐述了如何利用等比数列性质和逆元概念来高效求解此问题。

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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 $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$ .

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

Note that the modulo is unusual!

Input

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

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
inputCopy
2 2 3 3
+-+
outputCopy
7
inputCopy
4 1 5 1
-
outputCopy
999999228

Note

In the first example:

$(n \sum i = 0 s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7

In the second example:

$(n \sum i = 0 s_{i} a^{n - i} b^{i}) = - 1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = - 781 \equiv 999999228 (mod 10^{9} + 9)$

思路：因为每一项的正负是循环的，循环节的大小是k，并且保证了（n+1）是k的倍数。并且抛开正负不看的话这是一个等比数列，加上正负的话，因为正负是循环有规律的原因，我们可以把式子等分，每一份为k项，当成一个新的等比数列，k项为新等比数列中的一项，等比数列求和公式即可解出答案。

但是要注意还有取模的操作，因为我们在计算的时候会涉及公比（b/a）^k，而当a>b的时候取模就会爆精度。在这里就要通过求a的逆元来化除为乘。如果对逆元不了解的话，可以移步https://blog.youkuaiyun.com/baymax520/article/details/79997102

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int mod=1e9+9;
long long quickpow(long long a,long long b)//快速幂
{
    long long ans=1;
    a%=mod;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}

long long inv(long long a)//费马小定理求逆元
{
    return quickpow(a,mod-2);
}
int main()
{
    char s[100005];
    int n,a,b,k,i;
    while(scanf("%d %d %d %d",&n,&a,&b,&k)!=EOF)
    {
        scanf("%s",s);
        long long sum=0;
        for(i=0;i<k;i++)//先求出前k项的和
        {
            if(s[i]=='+')
                sum=(sum+quickpow(a,n-i)*quickpow(b,i)%mod+mod)%mod;
            else
                sum=(sum-quickpow(a,n-i)*quickpow(b,i)%mod+mod)%mod;
        }
        long long t=quickpow(b,k)*inv(quickpow(a,k))%mod;//使用逆元求公比(b/a)^k
        if(t==1)//公比为1和不为1的时候计算不同
        {
            sum=(sum*(n+1)/k)%mod;
            printf("%I64d\n",(sum+mod)%mod);
        }

        else
        {
           sum=sum*(quickpow(t,(n+1)/k)-1)%mod*inv(t-1)%mod;
            printf("%I64d\n",sum);
        }
    }
    return 0;
}