[Sdoi2017]数字表格 [莫比乌斯反演]

[Sdoi2017]数字表格

题意：求

\[\prod_{i=1}^n \prod_{j=1}^m f[(i,j)] \]

---

考场60分

其实多推一步就推倒了...

因为是乘，我们可以放到幂上

\[\prod_{d=1}^n \prod_{i=1}^{\frac{n}{d}}\prod_{i=1}^{\frac{m}{d}} f[d]^{[(i,j)=1]} \]

套路一直推完

\[\prod_{D=1}^n \prod_{d|D} f[d]^{\mu(\frac{D}{d}) \cdot \frac{n}{D} \cdot \frac{m}{D}} \]

用枚举倍数的方法预处理

\[\prod_{d|D} f[d]^{\mu(\frac{D}{d})} \]

的前缀积就行了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N=1e6+5, mo=1e9+7;
typedef long long ll;
inline int read() {
	char c=getchar(); int x=0, f=1;
	while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
	while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
	return x*f;
}

int Q, n, m, f[N]; ll g[N];
inline ll Pow(ll a, int b) {
	ll ans=1;
	for(; b; b>>=1, a=a*a%mo)
		if(b&1) ans=ans*a%mo;
	return ans;
}
int p[N], notp[N]; ll mu[N];
inline void mod(int &x) {if(x>=mo) x-=mo;}
inline void mod(ll &x) {if(x>=mo) x-=mo;}

void sieve(int n) {
	mu[1]=1;
	for(int i=2; i<=n; i++) { //printf("i %d\n",i);
		if(!notp[i]) p[++p[0]]=i, mu[i]=-1;
		for(int j=1; j<=p[0] && i*p[j]<=n; j++) { //printf("j %d\n",p[j]);
			notp[ i*p[j] ]=1;
			if(i % p[j] == 0) {
				mu[ i*p[j] ]=0;
				break;
			}
			mu[ i*p[j] ] = -mu[i];
		}
	}

	f[0]=0; f[1]=1; g[1]=1; g[0]=1;
	for(int i=2; i<=n; i++) mod(f[i] += f[i-1] + f[i-2]), g[i] = 1;
	for(int i=1; i<=n; i++) {
		ll inv = Pow(f[i], mo-2);
		for(int j=i, k=1; j<=n; j+=i, k++) if(mu[k]) ( g[j] *= (mu[k]==1 ? f[i] : inv) ) %= mo;
		(g[i] *= g[i-1]) %= mo;
	}
	//for(int i=1; i<=10; i++) printf("fg %d  %d  %lld\n", i, f[i], g[i]);
}

void cal(int n, int m) {
	if(n>m) swap(n, m);
	ll ans=1; int r;
	for(int i=1; i<=n; i=r+1) {
		r = min(n/(n/i), m/(m/i));
		( ans *= Pow(g[r] * Pow(g[i-1], mo-2) %mo, (ll)(n/i) * (m/i) % (mo-1)) ) %= mo;
	}
	printf("%lld\n", ans);
}
int main() {
	freopen("product.in", "r", stdin);
	freopen("product.out", "w", stdout);
	sieve(N-1);
	Q=read();
	for(int i=1; i<=Q; i++) {
		n=read(); m=read();
		cal(n, m);
	}
}