BZOJ2693: jzptab

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 10000000+5
 26 
 27 #define maxm 10000000
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 44 
 45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
 46 
 47 #define mod 100000009
 48 
 49 using namespace std;
 50 
 51 inline int read()
 52 
 53 {
 54 
 55     int x=0,f=1;char ch=getchar();
 56 
 57     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 58 
 59     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 60 
 61     return x*f;
 62 
 63 }
 64 int tot,p[maxn],mu[maxn];
 65 bool v[maxn];
 66 void get()
 67 {
 68     mu[1]=1;
 69     for2(i,2,maxm)
 70     {
 71         if(!v[i])p[++tot]=i,mu[i]=(i-(ll)i*i)%mod;
 72         for1(j,tot)
 73         {
 74             int k=i*p[j];
 75             if(k>maxm)break;
 76             v[k]=1;
 77             if(i%p[j])mu[k]=(ll)mu[p[j]]*mu[i]%mod;
 78             else {mu[k]=(ll)p[j]*mu[i]%mod;break;}
 79         }
 80     }
 81     for1(i,maxm)mu[i]=(mu[i]+mu[i-1])%mod;
 82 }
 83 inline int sum(int n,int m)
 84 {
 85     return ((ll)n*(n+1)/2%mod)*((ll)m*(m+1)/2%mod)%mod;
 86 }
 87 
 88 int main()
 89 
 90 {
 91 
 92     freopen("input.txt","r",stdin);
 93 
 94     freopen("output.txt","w",stdout);
 95     get();
 96 
 97     int T=read();
 98     while(T--)
 99     {
100         int n=read(),m=read(),ans=0;if(n>m)swap(n,m);
101         for(int i=1,j;i<=n;i=j+1)
102         {
103             j=min(n/(n/i),m/(m/i));
104             ans=(ans+(ll)(mu[j]-mu[i-1])*sum(n/i,m/i)%mod)%mod;
105         }
106         printf("%d\n",(ans+mod)%mod);
107     }
108 
109     return 0;
110 
111 }