P3704 [SDOI2017]数字表格

题目

P3704 [SDOI2017]数字表格

总算遇到一题不毒瘤的山东省选题了

做法

(prodlimits_{i=1}^nprodlimits_{j=1}^mf[gcd(i,j)])

根据套路枚举(gcd)

(prodlimits_{g=1}^{min(n,m)}prodlimits_{i=1}^nprodlimits_{j=1}[gcd(i,j)=d]f[d])
(prodlimits_{g=1}^{min(n,m)}f[g]^{ sumlimits_{i=1}^{frac{n}{d}} sumlimits_{j=1}^{frac{m}{d}} [gcd(i,j)=1] })

反演(sumlimits_{i=1}^{frac{n}{d}} sumlimits_{j=1}^{frac{m}{d}} [gcd(i,j)=1])：(sumlimits_{i=1}^{frac{n}{d}}mu(i)lfloor frac{n}{id} floor lfloor frac{m}{id} floor)

(prodlimits_{d=1}^{min(n,m)}f[d]^{ sumlimits_{i=1}^{frac{n}{d}}mu(i)lfloor frac{n}{id} floor lfloor frac{m}{id} floor })

枚举(T=id)：(prod_{T=1}^{min(n,m)}(prod_{d|T}f[d]^{mu(T/d)})^{[n/T][m/T]})

My complete code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL p=1e9+7;
const int maxn=10000000;
inline int Read(){
    int x(0),f(1); char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1; c=getchar();
    }
    while(c>='0'&&c<='9')
        x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
inline LL Pow(LL base,LL b){
    LL ret(1);
    while(b){
        if(b&1)
            ret=ret*base%p;
        base=base*base%p;
        b>>=1;
    }
    return ret;
}
int mu[maxn],prime[maxn];
LL f[maxn],g[maxn],F[maxn],sum[maxn];
bool visit[maxn];
inline void F_phi(int N){
    mu[1]=1;
    int tot(0);
    f[1]=1;
    for(int i=2;i<=N;++i){
        f[i]=(f[i-1]+f[i-2])%p,
        g[i]=Pow(f[i],p-2),
        F[i]=1;
        if(!visit[i]){
            prime[++tot]=i,
            mu[i]=-1;
        }
        for(int j=1;j<=tot&&i*prime[j]<=N;++j){
            visit[i*prime[j]]=true;
            if(i%prime[j]==0)
                break;
            else
                mu[i*prime[j]]=-mu[i];
        }
    }
    g[1]=F[1]=F[0]=1;
    for(int i=1;i<=N;++i){
        if(!mu[i])
            continue;
        for(int d=1;d*i<=N;++d)
            F[i*d]=1ll*F[i*d]*(mu[i]==1?f[d]:g[d])%p;
    }
    sum[0]=1ll;
    for(int i=1;i<=N;++i)
        sum[i]=sum[i-1]*F[i]%p;
}
int T;
int main(){
    F_phi(1000000);
    T=Read();
    while(T--){
    	int n(Read()),m(Read());
    	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));
    		LL ret=sum[r]*Pow(sum[l-1],p-2)%p;
    		ans=(ans*Pow(ret,1ll*(n/l)*(m/l)%(p-1)))%p;
        }
        printf("%lld
",ans);
    }
    return 0;
}/*
3
10000 100000
999 555
32465 485645
*/