51nod1227 平均最小公倍數

標籤：name   target   vector   sign   optimize   comment   sync   can   最小公倍數   

題意:求\(F(b)-F(a-1),F(n)=\sum_{i=1}^n\sum_{j=1}^i\frac{i}{(i,j)}\)
題解:先枚舉gcd,\(=\sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^i j[gcd(i,j)=1]=\sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\frac{i*\phi(i)}{2}+[i==1]\)
\(=\frac{1}{2}*(\sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i*\phi(i))+n\)
設f(x)=x*\phi(x),g(x)=x,g和f狄利克雷卷積,再杜教篩即可

//#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;ll phi[N],f[N],inv2=qp(2,mod-2),inv6=qp(6,mod-2);map<ll,ll>ff;bool mark[N];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(ll i=1;i<N;i++)    {        f[i]=phi[i]*i%mod;        add(f[i],f[i-1]);    }}ll getf(ll n){    if(n<N)return f[n];    if(ff.find(n)!=ff.end())return ff[n];    ll ans=n%mod*(n%mod+1)%mod*(n*2%mod+1)%mod*inv6%mod;    for(ll i=2,j;i<=n;i=j+1)    {        j=n/(n/i);        ll te=(j-i+1)%mod*((i+j)%mod)%mod*inv2%mod;        sub(ans,te*getf(n/i)%mod);    }    return ff[n]=ans;}ll cal(ll n){    ll ans=0;    for(ll i=1,j;i<=n;i=j+1)    {        j=n/(n/i);        add(ans,(j-i+1)*getf(n/i)%mod);    }    add(ans,n);    ans=ans*inv2%mod;    return ans;}int main(){    init();    ll a,b;    scanf("%lld%lld",&a,&b);    printf("%lld\n",((cal(b)-cal(a-1))%mod+mod)%mod);    return 0;}/****************************************/

51nod1227 平均最小公倍數