[SDOI2017]数字表格 [莫比乌斯反演]

​​传送门​​

反演模板吧...

到最后一步就是

然后预处理f(k) * mu(T/k)的前缀和与inv, 整除分块就可以了

#include<bits/stdc++.h>
#define N 1000050
#define Mod 1000000007
#define LL long long
using namespace std;
int T;
int prim[N], isp[N], tot, mu[N];
LL f[N], g[N], val[N], inv[N];
LL power(LL a,LL b){
  LL ans = 1;
  for(;b;b>>=1){
    if(b&1) ans = (ans*a) % Mod;
    a = (a*a) % Mod;
  } return ans;
}
void prework(){
  mu[1] = val[1] = 1;
  for(int i=2;i<=N-50;i++){
    val[i] = 1;
    if(!isp[i]) prim[++tot] = i, mu[i] = -1;
    for(int j=1;j<=tot;j++){
      if(prim[j] * i > N - 50) break;
      isp[prim[j] * i] = 1;
      if(i % prim[j] == 0) break;
      mu[i * prim[j]] = -mu[i];
    }
  }
  f[1] = f[2] = g[1] = g[2] = 1;
  for(int i=3;i<=N-50;i++){
    f[i] = (f[i-1] + f[i-2]) % Mod;
    g[i] = power(f[i], Mod-2);
  }
  for(int i=1;i<=N-50;i++){
    if(mu[i] == 0) continue;
    for(int j=i;j<=N-50;j+=i){
      val[j] = (val[j] * (mu[i] == 1 ? f[j/i] : g[j/i])) % Mod;
    }
  }
  for(int i=2;i<=N-50;i++) val[i] = (val[i] * val[i-1]) % Mod;
  inv[0] = 1; for(int i=1;i<=N-50;i++) inv[i] = power(val[i], Mod-2);
}
void Solve(){
  int n,m; scanf("%d%d",&n,&m);
  if(n>m) swap(n,m);
  LL ans = 1;
  for(int l=1,r;l<=n;l=r+1){
    int v1 = n/l, v2 = m/l;
    r = min(n/v1, m/v2);
    ans = (ans * power(val[r] * inv[l-1] % Mod, (LL)v1*(LL)v2)) % Mod;
  } printf("%lld\n",ans);
}
int main(){
  prework(); scanf("%d",&T);
  while(T--) Solve(); return 0;
}