BZOJ 2694 Lcm [莫比乌斯反演]

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然后将mu不为0的(没有平方因子)的h按调和级数枚举倍数算贡献就可以了

---

#include<bits/stdc++.h>
#define N 4000050
#define LL long long
#define P (1<<30)
using namespace std;
int T;
int prim[N],isp[N],tot,mu[N];
LL val[N];
void prework(){
  mu[1] = 1;
  for(int i=2;i<=N-50;i++){
    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]; 
    }
  }
  for(int i=1;i<=N-50;i++){
    if(mu[i] != 0){
      for(int j=i;j<=N-50;j+=i){
        LL tmp1 = (LL)i * (LL)(j/i); tmp1 %= P;
        LL tmp2 = (LL)(j/i) * (LL)mu[j/i]; tmp2 %= P;
        val[j] += (tmp1 * tmp2) % P; val[j] %= P;
      }
    }
  }
  for(int i=2;i<=N-50;i++) val[i] += val[i-1], val[i] %= P;
}
LL Sum(int x,int y){
  LL v1 = (LL)x * (LL)(x+1) / 2; v1 %= P;
  LL v2 = (LL)y * (LL)(y+1) / 2; v2 %= P;
  return (v1 * v2) % P;
}
void Solve(){
  int n,m; scanf("%d%d",&n,&m);
  if(n>m) swap(n,m);
  LL ans = 0;
  for(int l=1,r;l<=n;l=r+1){
    int v1 = n/l, v2 = m/l;
    r = min(n/v1, m/v2);
    ans += (Sum(v1,v2) * (val[r] - val[l-1])) % P;
    ans = (ans + P) % P;
  } printf("%lld\n",ans);
}
int main(){
  prework(); scanf("%d",&T);
  while(T--) Solve();
}