bzoj3561DZY Loves Math VI

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 #define mod 1000000007
 4 #define N 500001
 5 using namespace std;
 6 int n,m,cnt,mo[N],p[N>>1],vis[N];ll a[N],sum[N],ans;
 7 void predeal(){
 8     mo[1]=1;
 9     for(int i=2;i<N;i++){
10         if(!vis[i]){mo[i]=-1;p[++cnt]=i;}
11         for(int j=1;j<=cnt&&i*p[j]<N;++j){
12             vis[i*p[j]]=1;
13             if(i%p[j])mo[i*p[j]]=-mo[i];
14             else{mo[i*p[j]]=0;break;}
15         }
16     }
17 }
18 int quick(int a,int b){
19     int ret=1;
20     while(b){
21         if(b&1)ret=1ll*a*ret%mod;        
22         a=1ll*a*a%mod;b>>=1;
23     }
24     return ret;
25 }
26 int main(){
27     scanf("%d%d",&n,&m);predeal();if(n>m)swap(n,m);
28     for(int i=1;i<N;i++)a[i]=1;
29     for(int i=1;i<=n;i++){
30         ll res=0;
31         for(int j=1;j*i<=m;++j)
32         a[j]=a[j]*j%mod,sum[j]=(sum[j-1]+a[j])%mod;
33         for(int j=1;j*i<=n;++j)
34         if(mo[j])res=(res+mo[j]*a[j]*a[j]%mod*sum[n/i/j]%mod*sum[m/i/j]%mod)%mod;
35         ans=(ans+res*quick(i,i)%mod)%mod;
36     }
37     ans<0?ans+=mod:1;printf("%lld\n",ans);
38     return 0;
39 }