BZOJ 4816

Posted 2020-09-10 SiriusRen

思路：

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

$=\Pi_{d=1}^n\Pi_{i=1}^{\lfloor\frac{n}{d}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{d}\rfloor}f[d]*(gcd(i,j)==1)$

$\Sigma_{k=1}^n\Pi_{d|k}\Pi_{i=1}^{\lfloor\frac{n}{dk}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{dk}\rfloor}*f[d]*\mu(k)$

设dk=t

$=\Sigma_{t=1}^n\Pi_{i=1}^{\lfloor\frac{n}{t}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{t}\rfloor}\Pi_{d|t}f[d]*\mu(\frac{t}{d})$

$=\Sigma_{t=1}^n\Pi_{d|t}f[d]^{\mu(\frac{t}{d})*\lfloor\frac{n}{t}\rfloor*\lfloor\frac{m}{t}\rfloor}$

设$g(t)=\Pi_{d|t}f[d]^{\mu(\frac{t}{d})}$

$g(t)可以O(nlogn)预处理$

搞个前缀积

剩下的 喜闻乐见 分块

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1000005,mod=1000000007;
int n,m,cases,f[N],vis[N],prime[N],g[N],mu[N],tot;
typedef long long ll;
int pow(ll x,int y){
    ll res=1;
    while(y){
        if(y&1)res=res*x%mod;
        x=x*x%mod,y>>=1;
    }return res;
}
void shai(){
    mu[1]=f[1]=g[0]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])mu[i]=-1,prime[++tot]=i;
        for(int j=1;i*prime[j]<N&&j<=tot;j++){
            vis[i*prime[j]]=1,mu[i*prime[j]]=-mu[i];
            if(i%prime[j]==0){mu[i*prime[j]]=0;break;}
        }
        f[i]=(f[i-1]+f[i-2])%mod;
    }
    for(int i=1;i<N;i++)g[i]=1;
    for(int i=1;i<N;i++){
        int ni=pow(f[i],mod-2);
        for(int j=1;i*j<N;j++){
            if(mu[j]==1)g[i*j]=(1ll*g[i*j]*f[i])%mod;
            else if(mu[j]==-1)g[i*j]=(1ll*g[i*j]*ni)%mod;
        }
    }
    for(int i=1;i<N;i++)g[i]=(1ll*g[i]*g[i-1])%mod;
}
int main(){
    shai();
    scanf("%d",&cases);
    while(cases--){
        scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);
        ll ans=1;
        for(int l=1,r;l<=n;l=r+1){
            r=min(n/(n/l),m/(m/l));
            ans=ans*pow(1ll*g[r]*pow(g[l-1],mod-2)%mod,1ll*(n/l)*(m/l)%(mod-1))%mod;
        }printf("%lld\n",ans);
    }
}