【jzoj100006】【SDOI2017】【数字表格】【数论】

题目大意

解题思路

考虑生成一个新的序列g，使得f[i]=Πd|i(d!=i)g[d]

ans=Πni=1Πmj=1f[gcd(i,j)]

=Πni=1Πmj=1Πd|gcd(i,j)(d!=gcd(i,j))g[d]

=Πmin(n,m)i=1g[i](n/i)∗(m/i)

我们可以处理出g的前缀积，对于指数一样的i一起处理。

code

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define LF double#define LL long long#define ULL unsigned long long#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)#define fr(i,j,k) for(int i=begin[j][k];i;i=next[j][i])using namespace std;int const mn=1e6+9,mo=1e9+7,inf=1e9+7;int t,n,m;LL f[mn],g[mn],h[mn];LL Pow(LL x,LL y){    LL z=1;    while(y){        if(y&1)z=z*x%mo;        x=x*x%mo;        y>>=1;    }    return z;}int main(){    //freopen("product.in","r",stdin);    //freopen("product.out","w",stdout);    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d",&t);    int lim=1e6;    f[1]=g[0]=g[1]=h[0]=h[1]=1;    fo(i,2,lim)g[i]=f[i]=(f[i-2]+f[i-1])%mo;    fo(i,2,lim){        LL tmp=Pow(g[i],mo-2);        fo(j,2,lim/i)g[i*j]=g[i*j]*tmp%mo;        g[i]=g[i]*g[i-1]%mo;        h[i]=Pow(g[i],mo-2);    }    fo(cas,1,t){        scanf("%d%d",&n,&m);        if(n>m)swap(n,m);        LL ans=1;        for(int i=1,j;i<=n;i=j+1){            j=min(n/(n/i),m/(m/i));            ans=ans*Pow(g[j]*h[i-1]%mo,1ll*(n/i)*(m/i))%mo;        }        printf("%lld\n",ans);    }    return 0;}