題意 : 題目鏈接

分析 :

Tutorial 講的很清楚

至於為什麽這樣去考慮

算是一個經驗問題吧

如果一個問題要你給出模意義下的答案

就多考慮一下答案是要用逆元構造出來

也就說明有除法的存在

那麽可以去考慮等比數列或者等差數列求和公式等

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod = 1e9 + 9;

LL pow_mod(LL a, LL b)
{
    LL ret = 1;
    while(b){
        if(b & 1
 
) ret = (ret * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }return ret;
}

LL inv(LL a)
{ return pow_mod(a, mod-2); }

LL n, a, b, k, tmp;
int main(void)
{
    ios::sync_with_stdio(false); cin.tie(0);

    cin>>n>>a>>b>>k;

    string str;
    cin>>str;

    LL a0 
 
= 0;
    for(int i=0; i<k; i++){
        tmp = ( pow_mod(a, n-i) * pow_mod(b, i) ) % mod;
        if(str[i] == ‘-‘) a0 = ((a0 - tmp) + mod) % mod;
        else a0 = (a0 + tmp) % mod;
    }

    LL inv_a = inv(a);
    tmp = (b * inv_a)%mod;
    LL q = pow_mod(tmp, k);

    LL res;

    if(q == 1){
        res 
 
= (a0 * (n+1)/k)%mod;
        cout<<res<<endl;
        return 0;
    }

    LL qq = pow_mod(tmp, n+1);
    LL inv_q_1 = inv((q-1+mod)%mod);
    res = (a0 * (qq - 1 + mod)%mod )%mod;
    res = (res * inv_q_1) % mod;

    cout<<res<<endl;
    return 0;
}

Codeforces 963A Alternating Sum ( 思維 && 數論 )