技術標籤：題解

P5431 【模板】乘法逆元2

題意：

Solution：

這題主要是學一個想法。
s i = a 1 ∗ a 2 ∗ . . . ∗ a i 1 s i = 1 a 1 ∗ a 2 ∗ . . . ∗ a i 1 s i − 1 = 1 s i ∗ a i 1 a i = s i − 1 s i s_i=a_1*a_2*...*a_i \\ \frac{1}{s_i}=\frac{1}{a_1*a_2*...*a_i} \\ \frac{1}{s_{i-1}}=\frac{1}{s_i}*a_i \\ \frac{1}{a_i}=\frac{s_{i-1}}{s_i}

程式碼

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
char ch;
void read(int &x){
	x=
 
0;
	for(ch=getchar();ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar())
		x=(x<<3)+(x<<1)+(ch&15);
}
int n,p,k;
int a[5000005];
int ext_gcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x=1;y=0;
        return a;
    }
    int d=ext_gcd(b,a%
 
b,x,y);
    int tem=y;
    y=x-a/b*y;
    x=tem;
    return d;
}
int inv(int a,int b)
{
    int x,y;
    int d=ext_gcd(a,b,x,y);
    x=(x%p+p)%p;
    return x;
}
int Inv[5000005],b[5000005];
int main()
{
    read(n);read(p);read(k);
    int qaq=1;
    for(int i=0;i<n;i++)
    {
        read(a[i]);
        if(i==0)b[i]=a[i];
        else b[i]=1ll*b[i-1]*a[i]%p;
    }
    qaq=inv(b[n-1],p);
    for(int i=n-1;i>=0;i--)
    {
        if(i==0){Inv[i]=qaq;break;}
        Inv[i]=1ll*qaq*b[i-1]%p;
        qaq=1ll*qaq*a[i]%p;
    }
    int res=0;
    int q=1;
    for(int i=0;i<n;i++)
    {
        q=1ll*q*k%p;
        res=1ll*(res+1ll*Inv[i]*q%p)%p;
    }
    printf("%d",res);
    return 0;
}