51nod1237 of greatest common divisor and V3

Test instructions:\ (\SUM_{I=1}^N\SUM_{J=1}^NGCD (i,j) \)
Puzzle: First enumerate gcd,\ (\sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}[(I,j) =1]\)
Considering that the second half is the most common Möbius inversion, there \ (\sum_{d=1}^n\sum_{x=1}^{\lfloor \frac{n}{d} \rfloor}\mu (x) *{\lfloor \frac{n}{d*x} \ rfloor}^2\)
Enumeration dx=t,\ (=sum_{t=1}^n{\lfloor \frac{n}{t} \rfloor}^2\sum_{d|n}d*\mu (\frac{t}{d}) \)
Since \ (id=i*\phi\), inversion after \ (\phi=id*\mu\),\ (\sum_{t=1}^n{\lfloor \frac{n}{t} \rfloor}^2*\phi (t) \)
And finally, the sub-block + du screening phi

#pragma gcc optimize (2)//#pragma gcc optimize (3)//#pragma gcc optimize (4)//#pragma gcc optimize ("unroll-loops")//# pragma comment (linker, "/stack:200000000")//#pragma gcc optimize ("ofast,no-stack-protector")//#pragma gcc target (" Sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native ") #include <bits/stdc++.h> #define FI first#define SE Second#define DB Double#define mp make_pair#define pb push_back#define Pi ACOs ( -1.0) #define LL long Long#define VI vector& lt;int> #define MOD 1000000007#define ld long double#define C 0.5772156649#define ls l,m,rt<<1#define rs M+1,r,rt <<1|1#define PLL pair<ll,ll> #define PIL pair<int,ll> #define PLI pair<ll,int> #define PII pair <int,int>//#define CD complex<double> #define ULL unsigned long long#define base 1000000000000000000#define Max (a) (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) #define FIN freopen ("A.txt", "R", stdin) #define Fout freopen ("A.txt", "w", stdout) #define FIO Ios::sync_with_ Stdio (FALSE); Cin.tie (0) template<typename t>inline t const& MAX (T const &a,t const &AMP;B) {return a>b?a:b;} Template<typename t>inline T const& MIN (T const &a,t const &AMP;B) {return a<b?a:b;} inline void Add (ll &a,ll b) {a+=b;if (a>=mod) A-=mod;} inline void Sub (ll &a,ll b) {a-=b;if (a<0) A+=mod;} inline ll gcd (ll A,ll b) {return B?GCD (b,a%b): A;} inline ll QP (ll A,ll b) {ll ans=1;while (b) {if (b&1) ans=ans*a%mod;a=a*a%mod,b>>=1;} return ans;} inline ll QP (ll a,ll b,ll c) {ll ans=1;while (b) {if (b&1) ans=ans*a%c;a=a*a%c,b>>=1;} return ans;} Using namespace Std;const double eps=1e-8;const ll Inf=0x3f3f3f3f3f3f3f3f;const int n=5000000+10,maxn=3000000+10,inf=    0x3f3f3f3f;int prime[n],cnt;ll phi[n],inv2=qp (2,mod-2); bool Mark[n];map<ll,ll>phii;void init () {phi[1]=1;        for (int i=2;i<n;i++) {if (!mark[i]) prime[++cnt]=i,phi[i]=i-1;           for (int j=1;j<=cnt&&i*prime[j]<n;j++) {mark[i*prime[j]]=1; if (i%prime[j]==0) {phi[i*prime[j]]=phi[i]*prime[j];            Break        } phi[i*prime[j]]=phi[i]* (prime[j]-1);        }} for (int i=1;i<n;i++) {phi[i]%=mod;    Add (Phi[i],phi[i-1]);    }}ll Getphi (ll N) {if (n<n) return phi[n];    if (Phii.find (n)!=phii.end ()) return phii[n];    ll ans=n%mod* (n%mod+1)%mod*inv2%mod;        for (ll i=2,j;i<=n;i=j+1) {j=n/(n/i);    Sub (ans,1ll* (j-i+1)%mod*getphi (n/i)%mod); } return Phii[n]=ans;}    int main () {init ();    ll n,ans=0;scanf ("%lld", &n);        for (ll i=1,j;i<=n;i=j+1) {j=n/(n/i);        ll te= (n/i)%mod;te=te*te%mod;        ll P=getphi (j)-getphi (i-1);p = (p%mod+mod)%mod;    Add (Ans,te*p%mod);    } printf ("%lld\n", ans); return 0;} /****************************************/

51nod1237 Greatest common divisor and V3