loj #2000. 「SDOI2017」数字表格 （莫比乌斯）

题目链接：https://loj.ac/problem/2000

代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int MAXN=1e6+5;const int MOD=1e9+7;bool check[MAXN+10]; int prime[MAXN+10],f[MAXN][2],sum[MAXN],in[MAXN];int mu[MAXN+10];ll qpow(ll a,ll b){ll ans=1;a%=MOD;    for(ll i=b;i;i>>=1,a=a*a%MOD)        if(i&1)ans=ans*a%MOD;    return ans;}ll inv(ll x){return qpow(x,MOD-2);}void Moblus(){memset(check,false,sizeof(check));mu[1]=1;sum[1]=1;int tot=0;f[1][0]=1,f[0][0]=0;f[1][1]=1,f[0][1]=1;for(int i=2;i<MAXN;i++){f[i][0]=(f[i-1][0]+f[i-2][0])%MOD;f[i][1]=inv(f[i][0]);}for(int i=2;i<=MAXN;i++){sum[i]=1;if(!check[i]){prime[tot++]=i;mu[i]=-1;}for(int j=0;j<tot;j++){if(i*prime[j]>MAXN) break;check[i*prime[j]]=true;if(i%prime[j]==0){mu[i*prime[j]]=0;break;}else{mu[i*prime[j]]=-mu[i];}}}sum[0]=in[0]=1;for(int i=1;i<MAXN;i++){for(int j=i;j<MAXN;j+=i){if(mu[j/i]==-1)sum[j]=1LL*sum[j]*f[i][1]%MOD;if(mu[j/i]==1)sum[j]=1LL*sum[j]*f[i][0]%MOD;}}for(int i=1;i<MAXN;i++){sum[i]=1LL*sum[i-1]*sum[i]%MOD;in[i]=inv(sum[i]);}}ll cal(int n,int m){ll ret=1;int last;if(n<m)swap(n,m);for(int i=1;i<=m;i=last+1){last=min(n/(n/i),m/(m/i));ret=1LL*ret*qpow(1LL*sum[last]*in[i-1]%MOD,1LL*(n/i)*(m/i))%MOD;}return ret;}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);Moblus();int T;scanf("%d",&T);while(T--){int n,m;scanf("%d%d",&n,&m);printf("%lld\n",cal(n,m));}return 0;}