hdu5608杜教筛

标签：out   预处理   要求   a*   base   fir   complex   freopen   tac   

题意:给定函数\(f(x)\),有\(n^2-3*n+2=\sum_{d|n}f(d)\),求\(\sum_{i=1}^nf(i)\)
题解:很显然的杜教筛,假设\(g(n)=n^2-3*n+2\),那么有\(g=f*I\),由莫比乌斯反演,\(f=g*\mu\),可以O(nlogn)预处理到1e6,剩余部分杜教筛
我们先观察杜教筛的推导过程,假设要求\(s(n)=\sum_{i=1}^nf(i)\),
\(\sum_{i=1}^ng*f=\sum_{i=1}^n\sum_{d|i}g(d)f(\frac{i}{d})=\sum_{d=1}^ng(d)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}f(i)=\sum_{d=1}^ng(d)S(\lfloor \frac{n}{d} \rfloor)\)
\(S(n)=\sum_{i=1}^ng*f-\sum_{i=1}^ng(d)S(\lfloor \frac{n}{d} \rfloor)\)
我们考虑s就是我们要求的答案,g是常函数,那么I*f就是g,所以前半部分即\(\sum_{i=1}^ng(i)\)
分块处理后半部分,复杂度\(O(n^{\frac{2}{3})\)

//#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=1000000+10,maxn=3000000+10,inf=0x3f3f3f3f;

int prime[N],cnt,mu[N];
bool mark[N];
ll f[N];
map<ll,ll>ff;
map<ll,ll>::iterator it1;
ll inv3=qp(3,mod-2);
void init()
{
    mu[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-1;
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<N;i++)
        for(int j=i;j<N;j+=i)
        {
            ll te=1ll*(j/i-2)*(j/i-1)*mu[i];
            te=(te%mod+mod)%mod;
            add(f[j],te);
        }
//    printf("%lld\n",f[1000000]);
    for(int i=1;i<N;i++)add(f[i],f[i-1]);
}
ll getf(ll n)
{
    if(n<N)return f[n];
    if((it1=ff.find(n))!=ff.end())return it1->se;
    ll ans=n*(n+1)%mod*(n-4)%mod*inv3%mod+2ll*n%mod;
    ans=(ans%mod+mod)%mod;
    for(ll i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        sub(ans,1ll*(j-i+1)*getf(n/i)%mod);
    }
    return ff[n]=ans;
}
int main()
{
    init();
    int T;scanf("%d",&T);
    while(T--)
    {
        ll n;scanf("%lld",&n);
        printf("%lld\n",getf(n));
    }
    return 0;
}
/********************

********************/

hdu5608杜教筛

标签：out   预处理   要求   a*   base   fir   complex   freopen   tac   

原文地址：https://www.cnblogs.com/acjiumeng/p/9739793.html