【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;
}