P5431 模意义下的乘法逆元

P5431 模意义下的乘法逆元

思路

求前缀积，然后求第\(n\) 个数的逆元

\(sv[i - 1] = sv[i] * a[i]\)最后在根据 \(inv[i] = prefix[i - 1] * sv[i]\) 即可。

代码

#include <bits/stdc++.h>

using i64 = long long;

const int N = 5e6 + 10;

int n, p, k, arr[N], prefix[N], sv[N], K[N];

int fast_pow(int a, int b, int mod) {
	int res = 1;
	while (b) {
		if (b & 1) res = (i64) res * a % mod;
		a = (i64) a * a % mod;
		b >>= 1;
	}
	return res;
}

template<class T>
T read(T &x){
    x=0;char ch=0;
    while(ch < '0' || ch > '9')ch=getchar();
    while(ch >= '0' && ch <= '9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}

int main() {
	read(n), read(p), read(k);

	prefix[0] = K[0] = 1;

	for (int i = 1; i <= n; i ++) {
		read(arr[i]);
		prefix[i] = (i64) arr[i] * prefix[i - 1] % p;
		K[i] = (i64) K[i - 1] * k % p;	
	} 

	sv[n] = fast_pow(prefix[n], p - 2, p);

	i64 res = 0;


	for (int i = n; i >= 1; i --) {
		sv[i - 1] = (i64) sv[i] * arr[i] % p;
		res += (((i64) sv[i] * prefix[i - 1] % p) * K[i])% p;
		res %= p;
	}

	printf("%lld", res);
}