bzoj2693 莫比乌斯反演

Description

Hint

T <= 10000
N, M<=10000000

 

https://wenku.baidu.com/view/fbec9c63ba1aa8114431d9ac.html

popoqqq的论文

 

 #pragma GCC optimize(2)
 #pragma G++ optimzie(2)
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cstdio>
 #include<algorithm>

 #define mod 100000009
 #define N 10000007
 #define ll long long
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
     while(isdigit(ch)){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }

 int T,n,m;
 int tot;
 bool flag[N];
 ll h[N],pri[N];

 void init()
 {
     h[]=;
     for (int i=;i<N;i++)
     {
         if(!flag[i])
         {
             pri[++tot]=i;
             h[i]=(i-(ll)i*i)%mod;
         }
         for (int j=;j<=tot&&pri[j]*i<N;j++)
         {
             flag[pri[j]*i]=true;
             if(i%pri[j]==)
             {
                 h[pri[j]*i]=(pri[j]*h[i])%mod;
                 break;
             }
             else h[pri[j]*i]=(h[pri[j]]*h[i])%mod;
         }
     }
     for (int i=;i<N;i++)
         (h[i]+=h[i-])%=mod;
 }
 inline ll sum(ll x,ll y)
 {
     x%=mod,y%=mod;
     x*=(x+),(x=x/)%=mod;
     y*=(y+),(y=y/)%=mod;
     return x*y%mod;
 }
 ll query(int n,int m)
 {
     ll res=;
     if(n>m)swap(n,m);
     for (int i=,last;i<=n;i=last+)
     {
         last=min(n/(n/i),m/(m/i));
         res+=sum(n/i,m/i)*(h[last]-h[i-])%mod;
         res%=mod;
     }
     return (res%mod+mod)%mod;
 }
 int main()
 {
     T=read(),init();
     while(T--)
     {
         n=read(),m=read();
         printf("%lld\n",query(n,m));
     }
 }