• bzoj3944: Sum 杜教筛板子题


    板子题(卡常)
    也可能是用map太慢了

    /**************************************************************
        Problem: 3944
        User: walfy
        Language: C++
        Result: Accepted
        Time:9932 ms
        Memory:84304 kb
    ****************************************************************/
     
    //#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=400000+10,inf=0x3f3f3f3f;
     
    int prime[N],cnt,mu[N];
    ll phi[N];
    bool mark[N];
    map<int,int>muu;
    map<int,int>::iterator it1;
    map<int,ll>phii;
    map<int,ll>::iterator it2;
    void init()
    {
        mu[1]=phi[1]=1;
        for(int i=2;i<N;i++)
        {
            if(!mark[i])prime[++cnt]=i,mu[i]=-1,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)
                {
                    mu[i*prime[j]]=0;
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
        for(int i=1;i<N;i++)mu[i]+=mu[i-1],phi[i]+=phi[i-1];
    }
    ll getmu(ll n)
    {
        if(n<N)return mu[n];
        if((it1=muu.find(n))!=muu.end())return it1->se;
        ll ans=1;
        for(ll i=2,j;i<=n;i=j+1)
        {
            j=n/(n/i);
            ans-=1ll*(j-i+1)*getmu(n/i);
        }
        return muu[n]=ans;
    }
    ll getphi(ll n)
    {
        if(n<N)return phi[n];
        if((it2=phii.find(n))!=phii.end())return it2->se;
        ll ans=n*(n+1)/2;
        for(ll i=2,j;i<=n;i=j+1)
        {
            j=n/(n/i);
            ans-=1ll*(j-i+1)*getphi(n/i);
        }
        return phii[n]=ans;
    }
    int main()
    {
        init();
        int T;scanf("%d",&T);
        while(T--)
        {
            ll n;scanf("%lld",&n);
            printf("%lld %lld
    ",getphi(n),getmu(n));
        }
        return 0;
    }
    /********************
    10
    2147483638
    2147483639
    2147483640
    2147483641
    2147483642
    2147483643
    2147483644
    2147483645
    2147483646
    2147483647
    ********************/
    

    不用map的版本

    /**************************************************************
        Problem: 3944
        User: walfy
        Language: C++
        Result: Accepted
        Time:7600 ms
        Memory:67700 kb
    ****************************************************************/
     
    //#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=2000000+10,maxn=400000+10,inf=0x3f3f3f3f;
     
    int prime[N],cnt,mu[N];
    ll phi[N],n,muu[N],phii[N];
    bool mark[N],vis[N];
    void init()
    {
        mu[1]=phi[1]=1;
        for(int i=2;i<N;i++)
        {
            if(!mark[i])prime[++cnt]=i,mu[i]=-1,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)
                {
                    mu[i*prime[j]]=0;
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
        for(int i=1;i<N;i++)mu[i]+=mu[i-1],phi[i]+=phi[i-1];
    }
    void get(ll x)
    {
        if(x<N)return ;
        if(vis[n/x])return ;
        vis[n/x]=1;
        phii[n/x]=x*(x+1)/2;
        muu[n/x]=1;
        for(ll i=2,j;i<=x;i=j+1)
        {
            j=x/(x/i);
            if(x/i<N)phii[n/x]-=1ll*(j-i+1)*phi[x/i],muu[n/x]-=1ll*(j-i+1)*mu[x/i];
            else
            {
                get(x/i);
                phii[n/x]-=1ll*(j-i+1)*phii[n/(x/i)];
                muu[n/x]-=1ll*(j-i+1)*muu[n/(x/i)];
            }
        }
    }
    int main()
    {
        init();
        int T;scanf("%d",&T);
        while(T--)
        {
            memset(vis,0,sizeof vis);
            scanf("%lld",&n);
            if(n<N)printf("%lld %d
    ",phi[n],mu[n]);
            else
            {
                memset(vis,0,sizeof vis);get(n);
                printf("%lld %lld
    ",phii[1],muu[1]);
            }
        }
        return 0;
    }
    /********************
    10
    2147483638
    2147483639
    2147483640
    2147483641
    2147483642
    2147483643
    2147483644
    2147483645
    2147483646
    2147483647
    ********************/
    
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/9523233.html
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