P3768 简单的数学题 [莫比乌斯反演+杜教筛]

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用杜教筛筛就可以了(g 设为Id^2) 

---

#include<bits/stdc++.h>
#define N 10000050
#define LL long long
using namespace std;
LL P,n;
int prim[N],isp[N],tot; LL phi[N];
map<LL,LL> Phi;
LL inv2, inv6;
LL power(LL a, LL b){
  LL ans = 1;
  for(;b;b>>=1){ if(b&1) ans = (ans*a) % P; a = (a*a) % P;}
  return ans;
}
void prework(){
  inv2 = power(2, P-2);
  inv6 = power(6, P-2);
  phi[1] = 1;
  for(int i=2;i<=N-50;i++){
    if(!isp[i]) prim[++tot] = i, phi[i] = i-1;
    for(int j=1;j<=tot;j++){
      if(i * prim[j] > N - 50) break;
      isp[i * prim[j]] = 1;
      if(i % prim[j] == 0){
        phi[i * prim[j]] = phi[i] * prim[j];
        break;
      }
      phi[i * prim[j]] = phi[i] * (prim[j] - 1);
    }
  }
  for(int i=1;i<=N-50;i++) phi[i] = phi[i] % P * i % P * i % P;
  for(int i=2;i<=N-50;i++) phi[i] += phi[i-1], phi[i] %= P;
}
LL Sum(LL x){x %= P; return x * (x+1) % P * inv2 % P;}
LL G(LL x){ x %= P; return x * (x+1) % P * (x*2+1) % P * inv6 % P;}
LL getf(LL x){
  if(x<=N-50) return phi[x];
  if(Phi[x]) return Phi[x];
  LL ans = Sum(x); ans = (ans * ans) % P;
  for(LL l=2,r;l<=x;l=r+1){
    LL val = x/l; r = x/val;
    ans -= (G(r) - G(l-1)) % P * getf(val) % P;
    ans = (ans + P) % P; 
  } return Phi[x] = ans;
}
int main(){
  scanf("%lld%lld",&P,&n); prework();
  LL ans = 0;
  for(LL l=1,r;l<=n;l=r+1){
    LL val = n/l; r = n/val;
    LL tmp = Sum(val); tmp = (tmp * tmp) % P;
    ans += tmp * (getf(r) - getf(l-1)) % P;
    ans = (ans + P) % P;
  } printf("%lld",ans); 
}