「解题报告」P5431 乘法逆元 2

题目链接：【模板】乘法逆元 2

这道题不建议叫乘法逆元，可以直接当一道数学题去处理，我们观察这个式子 $\sum _{i=1}^n\frac{k^i}{a_i}$，那么我们直接通分就和即可，分子就是 $\sum _{i=1}^n{k}^i\cdot (a_1\sim a_i-1\cdot a_i+1\sim a_n)$，预处理出前缀积和后缀积即可，最后乘上 $1\sim n$ 阶乘的逆元即可。

这道题卡时间，用快读快写可以过。

点击查看代码

#define M 5000010
using namespace std;
int n, p, k, ans, base = 1;
int f[M], g[M], inv[M], a[M];
inline int read() {
	int f(1) , x(0);
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
	for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return f * x;
}
inline void write(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
int QuickPow(int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = res * a % p;
		a = a * a % p;
		b >>= 1;
	}
	return res;
}
int main() {
	n = read(), p = read(), k = read();
	for (int i = 1; i <= n; i++)
		a[i] = read();
	g[0] = f[n+1] = 1;
	for (int i = 1; i <= n; i++)
		g[i] = g[i-1] * a[i] % p;
	for (int i = n; i >= 1; i--)
		f[i] = f[i+1] * a[i] % p;
	for (int i = 1; i <= n; i++) {
		base = base * k % p;
		ans = (ans + base * g[i-1] % p * f[i+1] % p) % p;
	}
	int w = QuickPow(g[n], p - 2);
	write((ans * w) % p);
	return 0;
}