「UR#5」怎样跑得更快

「UR#5」怎样跑得更快

膜这个您就会了

下面是复读机mangoyang

我们要求

\[\sum_{j=1}^n \gcd(i,j)^{c-d} j^d x_j=\frac{b_i}{i^d}
\]

随便设一下

\[\sum_{j=1}^n f(\gcd(i,j))h(j)=g(i) \\
\sum_{d|i}\sum_{j=1}^n [\gcd(i,j)=d]f(d)h(j)=g(i) \\
\sum_{d|i}\sum_{d|j}f_r(d)h(j)=g(i)
\]

这里用到了第一个莫比乌斯反演，已知 \(f(d)\) 求出 \(f_r(d)\)

记 \(f_z(d)=\sum_{j=1}^n [d|j]h(j)\)

\[\sum_{d|i} f_r(d)f_z(d)=g(i)
\]

这里用第二个莫比乌斯反演，已知 \(g(i)\) 求出 \(f_r(d)f_z(d)\) ，除一下可以得到 \(f_z(d)\) 。

最后用第三个莫比乌斯反演，已知 \(f_z(d)\) 求出 \(h(j)\) 即可。

code

/*program by mangoyang*/
#pragma GCC optimize("Ofast", "inline")
#include<bits/stdc++.h>
#define inf (0x3f3f3f3f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
	int ch = 0, f = 0; x = 0;
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
	for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
	if(f) x = -x;
}
const int N = 114514, mod = 998244353;
int a[N], b[N], inv1[N], inv2[N], n, q, c, d;
inline void up(int &x, int y){
	x = x + y >= mod ? x + y - mod : x + y;
}
inline int Pow(int a, int b){
	int ans = 1;
	b = (b % (mod - 1) + mod - 1) % (mod - 1);
	for(; b; b >>= 1, a = 1ll * a * a % mod)
		if(b & 1) ans = 1ll * ans * a % mod;
	return ans;
}
inline void gao1(int *a){
	for(int i = 1; i <= n; i++)
		for(int j = i + i; j <= n; j += i)
			up(a[j], mod - a[i]);
}
inline void gao2(int *a){
	for(int i = n; i >= 1; i--)
		for(int j = i + i; j <= n; j += i)
			up(a[i], mod - a[j]);
}
int main(){
	read(n), read(c), read(d), read(q);
	for(int i = 1; i <= n; i++)
		inv1[i] = Pow(i, c - d);
	for(int i = 1; i <= n; i++)
		inv2[i] = Pow(i, -d);
	gao1(inv1);
	for(int i = 1; i <= n; i++)
		inv1[i] = Pow(inv1[i], -1);
	while(q--){
		for(int i = 1; i <= n; i++){
			read(b[i]);
			b[i] = 1ll * b[i] * inv2[i] % mod;
		}
		gao1(b);
		int flag = 0;
		for(int i = 1; i <= n; i++)
			if(!inv1[i] && b[i]){
				flag = 1; break;
			}
			else b[i] = 1ll * b[i] * inv1[i] % mod;
		if(flag){
			puts("-1");
			continue;
		}
		gao2(b);
		for(int i = 1; i <= n; i++)
			b[i] = 1ll * b[i] * inv2[i] % mod;
		for(int i = 1; i <= n; i++)
			printf("%d ", b[i]);
		puts("");
	}
	return 0;
}