【UR #5】怎样跑得更快 莫比乌斯反演

版权声明：本文是蒟蒻写的，转载。。。随便吧 https://blog.csdn.net/xgc_woker/article/details/86651106

Description
给你一个b序列，求x序列，b和x满足下式。
$\sum_{j = 1}^{n} gcd(i, j)^c \cdot lcm(i, j)^d \cdot x_j \equiv b_i \pmod{p}$

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Sample Input
3 1 0 2
1 0 0
1 2 3

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Sample Output
499122179 998244352 499122176
998244352 1 1

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式子首先可以化成：
$\sum_{j = 1}^{n} j^dgcd(i, j)^{c-d} \cdot x_j \equiv \frac {b_i}{i^d} \pmod{p}$
我们考虑设 $g_i$为 $i^d$， $f_i$为 $i^{c-d}$， $w_i$为 $\frac {b_i}{i^d}$。
$\sum_{j = 1}^{n} g_j \cdot f_{gcd(i, j)} \cdot x_j \equiv w_i \pmod{p}$
考虑莫反搞掉这个 $gcd$，设 $fr_i=\sum_{d|i}f_d \cdot \mu_{\frac id}$。
设 $t_i=x_i \cdot g_i$，那么式子变为：
$\sum_{j = 1}^{n}\sum_{d|i,d|j} fr_d \cdot t_j \equiv w_i \pmod{p}$
$\sum_{d|i} fr_d \sum_{d|j} t_j \equiv w_i \pmod{p}$
设 $q_d=\sum_{d|j} t_j$，
$\sum_{d|i} fr_d\cdot q_d \equiv w_i \pmod{p}$
设 $F_d=fr_d\cdot q_d$，
莫反一下，这样就搞得出 $F$，进而搞得出 $q$。
对于这个 $q$再莫反一下，就搞得出 $t$。
搞出了 $t$，这提就做完了。

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#include <cstdio>
#include <cstring>

using namespace std;
typedef long long LL;
const int N = 300001;
const LL mod = 998244353;
int read() {
	int s = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;
}
void put(int x) {
	if(x >= 10) put(x / 10);
	putchar(x % 10 + '0');
}

LL w[N], g[N], f[N], fr[N], F[N], q[N], t[N], x[N];
int plen, p[N], mu[N];
bool v[N];

void get_p() {
	mu[1] = 1;
	for(int i = 2; i < N; i++) {
		if(!v[i]) p[++plen] = i, mu[i] = -1;
		for(int j = 1; j <= plen && (LL)p[j] * i < N; j++) {
			v[i * p[j]] = 1;
			if(i % p[j] == 0) {mu[i * p[j]] = 0; break;}
			mu[i * p[j]] = -mu[i];
		}
	}
}

LL pow_mod(LL a, int k) {
	LL ans = 1;
	k = (k + mod - 1) % (mod - 1);
	while(k) {
		if(k & 1) (ans *= a) %= mod;
		(a *= a) %= mod; k /= 2;
	} return ans;
}

int main() {
	int n = read(), c = read(), d = read(), tt = read();
	get_p();
	for(int i = 1; i <= n; i++) g[i] = pow_mod(i, d), f[i] = pow_mod(i, c - d);
	for(int i = 1; i <= n; i++) {
		for(int j = i, s = 1; j <= n; j += i, s++) {
			(fr[j] += mu[s] * f[i]) %= mod;
		} (fr[i] += mod) %= mod;
	}
	while(tt--) {
		bool bk = 0;
		memset(F, 0, sizeof(F));
		memset(t, 0, sizeof(t));
		for(int i = 1; i <= n; i++) w[i] = read(), w[i] = w[i] * pow_mod(g[i], mod - 2) % mod;
		for(int i = 1; i <= n; i++) {
			for(int j = i, s = 1; j <= n; j += i, s++) {
				(F[j] += mu[s] * w[i]) %= mod;
			} (F[i] += mod) %= mod;
		} for(int i = 1; i <= n; i++) {
			if(F[i] != 0 && fr[i] == 0) {bk = 1; break;}
			q[i] = F[i] * pow_mod(fr[i], mod - 2) % mod;
		} if(bk) {puts("-1"); continue;}
		for(int i = 1; i <= n; i++) {
			for(int j = i, s = 1; j <= n; j += i, s++) {
				(t[i] += mu[s] * q[j]) %= mod;
			} (t[i] += mod) %= mod;
		} for(int i = 1; i <= n; i++) {
			if(t[i] != 0 && g[i] == 0) {bk = 1; break;}
			x[i] = t[i] * pow_mod(g[i], mod - 2) % mod;
		} if(bk) {puts("-1"); continue;}
		for(int i = 1; i <= n; i++) put(x[i]), putchar(' ');
		putchar('\n');
	}
	return 0;
}