Codeforces 963A Alternating Sum ( 思维 && 数论 )

Posted 2020-10-31 rubbishes

题意 : 题目链接

 

分析 :

Tutorial 讲的很清楚

 

至于为什么这样去考虑

算是一个经验问题吧

如果一个问题要你给出模意义下的答案

就多考虑一下答案是要用逆元构造出来

也就说明有除法的存在

那么可以去考虑等比数列或者等差数列求和公式等

 

#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;
}

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