「Luogu-P5431」【模板】乘法逆元2

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字数统计: 40 | 阅读时长 ≈ 1

给定$n$个正整数$a_i$，求出在$\bmod p$意义下$\sum\limits_{i=1}^{n}k^i\cdot a_i^{-1}$

$\texttt{Data Range:}1\leq n\leq 5\times 10^6,2\leq k

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题解

代码

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#include<bits/stdc++.h>
using namespace std;
typedef int ll;
typedef long long int li;
const ll MAXN=5e6+51;
char buf[1<<15],*st=buf,*ed=buf;
ll cnt,mod,k,res,cur;
ll prefix[MAXN],invp[MAXN],num[MAXN];
inline char gcx()
{
    return st==ed&&(ed=(st=buf)+fread(buf,1,32768,stdin),ed==st)?0:*st++; 
}
inline ll read()
{
    register ll num=0,neg=1;
    register char ch=gcx();
    while(!isdigit(ch)&&ch!='-')
    {
        ch=gcx();
    }
    if(ch=='-')
    {
        neg=-1;
        ch=gcx();
    }
    while(isdigit(ch))
    {
        num=(num<<3)+(num<<1)+(ch-'0');
        ch=gcx();
    }
    return num*neg;
} 
inline ll qpow(ll base,ll exponent)
{
    ll res=1;
    while(exponent)
    {
        if(exponent&1)
        {
            res=(li)res*base%mod;
        }
        base=(li)base*base%mod,exponent>>=1;
    }
    return res;
} 
int main()
{
    cnt=read(),mod=read(),k=read(),prefix[0]=1;
    for(register int i=1;i<=cnt;i++)
    {
        prefix[i]=1ll*prefix[i-1]*(num[i]=read())%mod;
    }
    invp[cnt]=qpow(prefix[cnt],mod-2);
    for(register int i=cnt-1;i;i--)
    {
        invp[i]=1ll*invp[i+1]*num[i+1]%mod;
    }
    res=(res+1ll*k*invp[1]%mod)%mod,cur=1ll*k*k%mod;
    for(register int i=2;i<=cnt;i++)
    {
        res=(res+1ll*cur*invp[i]%mod*prefix[i-1]%mod)%mod;
        cur=1ll*cur*k%mod;
    }
    printf("%d\n",res);
}