bzoj2693 莫比乌斯反演

Description

Hint

T <= 10000
N, M<=10000000

 
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 1 #pragma GCC optimize(2)
 2 #pragma G++ optimzie(2)
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<algorithm>
 8 
 9 #define mod 100000009
10 #define N 10000007
11 #define ll long long
12 using namespace std;
13 inline int read()
14 {
15     int x=0,f=1;char ch=getchar();
16     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
17     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
18     return x*f;
19 }
20 
21 int T,n,m;
22 int tot;
23 bool flag[N];
24 ll h[N],pri[N];
25 
26 void init()
27 {
28     h[1]=1;
29     for (int i=2;i<N;i++)
30     {
31         if(!flag[i])
32         {
33             pri[++tot]=i;
34             h[i]=(i-(ll)i*i)%mod;
35         }
36         for (int j=1;j<=tot&&pri[j]*i<N;j++)
37         {
38             flag[pri[j]*i]=true;
39             if(i%pri[j]==0)
40             {
41                 h[pri[j]*i]=(pri[j]*h[i])%mod;
42                 break;
43             }
44             else h[pri[j]*i]=(h[pri[j]]*h[i])%mod;
45         }
46     }
47     for (int i=1;i<N;i++)
48         (h[i]+=h[i-1])%=mod;
49 }
50 inline ll sum(ll x,ll y)
51 {
52     x%=mod,y%=mod;
53     x*=(x+1),(x=x/2)%=mod;
54     y*=(y+1),(y=y/2)%=mod;
55     return x*y%mod;
56 }
57 ll query(int n,int m)
58 {
59     ll res=0;
60     if(n>m)swap(n,m);
61     for (int i=1,last;i<=n;i=last+1)
62     {
63         last=min(n/(n/i),m/(m/i));
64         res+=sum(n/i,m/i)*(h[last]-h[i-1])%mod;
65         res%=mod;
66     }
67     return (res%mod+mod)%mod;
68 }
69 int main()
70 {
71     T=read(),init();
72     while(T--)
73     {
74         n=read(),m=read();
75         printf("%lld\n",query(n,m));
76     }
77 }

 

posted @ 2018-03-08 20:45  Kaiser-  阅读(125)  评论(0编辑  收藏  举报