[Euler function] [convolution] Luogu P3768 simple math

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Enter an integer n and an integer p, you need to find (Σi = 1nΣj = 1nijgcd (i , j)) mod p (\ sum_ {i = 1} ^ n \ sum_ {j = 1} ^ n ijgcd (I, J)) ~ MOD ~ P ( [Sigma I = . 1 n- [Sigma J = . 1 n- I J G C D ( I , J ) ) m O D P, where gcd (a, b) denotes a and b the greatest common divisor.  

 

answer

 

Code

 1 #include <map>
 2 #include <cstdio>
 3 #include <iostream>
 4 #define N 4600000
 5 #define sqr(x) (x)*(x)
 6 #define ll long long
 7 using namespace std;
 8 ll p,n,inv,pos,r,phi[N],q[N],bz[N],s[N];
 9 map<ll,ll> m;
10 ll calc1(ll x) { x%=p; return x*(x+1)%p*(2*x+1)%p*inv%p; }
11 ll calc2(ll x) { x%=p; return sqr((x*(x+1)/2)%p)%p; }
12 ll ksm(ll a,ll b) { ll r=1; for (;b;b>>=1,a=a*a%p) if (b&1) r=r*a%p; return r; }
13 ll calc(ll x)
14 {
15     if (x<N) return s[x];
16     if (m[x]) return m[x];
17     ll pos,r=calc2(x);
18     for (ll i=2;i<=x;i=pos+1) pos=x/(x/i),r=(r-(calc1(pos)-calc1(i-1)+p)%p*calc(x/i)%p)%p;
19     m[x]=r=(r+p)%p; return r;
20 }
21 int main()
22 {
23     scanf("%lld%lld",&p,&n),inv=ksm(6,p-2),phi[1]=1;
24     for (ll i=2;i<N;i++)
25     {
26         if (!bz[i]) q[++q[0]]=i,phi[i]=i-1;
27         for (ll j=1;j<=q[0];j++)
28             if (i*q[j]<N)
29             {
30                 bz[i*q[j]]=1;
31                 if (i%q[j]==0) { phi[i*q[j]]=phi[i]*q[j]; break; }
32                 phi[i*q[j]]=phi[i]*(q[j]-1);
33             }
34             else break;
35     }
36     for (ll i=1;i<N;i++) s[i]=(s[i-1]+i*i%p*phi[i]%p)%p;
37     for (ll i=1;i<=n;i=pos+1) pos=n/(n/i),r=(r+(calc(pos)-calc(i-1)+p)%p*calc2(n/i)%p)%p;
38     printf("%lld",r=(r+p)%p); 
39 }