51nod1238 最小公倍數之和 V3
阿新 • • 發佈:2018-10-04
clas i++ ble optimize opc 51nod bit return n+1
\(=\sum_{d=1}^nd2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i*\frac{i*\phi(i)+[i==1]}{2}-1\)
\(=\sum_{d=1}^nd\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i^2*\phi(i)\)
假設\(f(i)=i^2*\phi(i),S(n)=\sum_{i=1}^nf(i)\)
由杜教篩\(g(1)S(n)=\sum_{i=1}^ng*f-\sum_{d=2}g(d)S(\lfloor \frac{n}{d} \rfloor)\)
考慮\(g(n)=n^2\),g和f狄利克雷卷積\(=\sum_{d|n}d^2\phi(d){\frac{n}{d}}^2=n^3\)
那麽\(S(n)=\sum_{i=1}^n n^3-\sum_{d=2}d^2S(\lfloor \frac{n}{d} \rfloor)\)
題意:求\(\sum_{i=1}^n\sum_{j=1}^n\frac{i*j}{gcd(i,j)}\)
題解:先枚舉gcd,\(\sum_{d=1}^nd\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}i*j[(i,j)==1]\)
\(=\sum_{d=1}^nd2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{i}i*j[(i,j)==1]-1\)
\(=\sum_{d=1}^nd2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i\sum_{j=1}^{i}j[(i,j)==1]-1\)
\(=\sum_{d=1}^nd2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i*\frac{i*\phi(i)+[i==1]}{2}-1\)
\(=\sum_{d=1}^nd\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i^2*\phi(i)\)
假設\(f(i)=i^2*\phi(i),S(n)=\sum_{i=1}^nf(i)\)
由杜教篩\(g(1)S(n)=\sum_{i=1}^ng*f-\sum_{d=2}g(d)S(\lfloor \frac{n}{d} \rfloor)\)
考慮\(g(n)=n^2\),g和f狄利克雷卷積\(=\sum_{d|n}d^2\phi(d){\frac{n}{d}}^2=n^3\)
那麽\(S(n)=\sum_{i=1}^n n^3-\sum_{d=2}d^2S(\lfloor \frac{n}{d} \rfloor)\)
//#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<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,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((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 &b){return a>b?a:b;} template<typename T> inline T const& MIN(T const &a,T const &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,phi[N]; bool mark[N]; ll f[N],inv2=qp(2,mod-2),inv6=qp(6,mod-2); 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++) { f[i]=1ll*i*i%mod*phi[i]%mod; add(f[i],f[i-1]); } } ll getf(ll n) { if(n<N)return f[n]; if(phii.find(n)!=phii.end())return phii[n]; ll ans=n%mod*((n+1)%mod)%mod*inv2%mod;ans=ans*ans%mod; for(ll i=2,j;i<=n;i=j+1) { j=n/(n/i); ll tj=j%mod,ti=i%mod; ll te=tj*(tj+1)%mod*(2ll*tj+1)%mod*inv6%mod-(ti-1)*ti%mod*(2ll*ti-1)%mod*inv6%mod; te=(te%mod+mod)%mod; sub(ans,te*getf(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); add(ans,(j-i+1)%mod*((i+j)%mod)%mod*inv2%mod*getf(n/i)%mod); } printf("%lld\n",ans); return 0; } /******************** ********************/
51nod1238 最小公倍數之和 V3