ACM-ICPC 2018 徐州赛区网络预赛 D. Easy Math

Easy Math

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•  1000ms
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Given a positive integers nn , Mobius function mu(n)μ(n) is defined as follows:

displaystyle mu(n) = egin{cases} 1 &n = 1 \ (-1)^k & n = p_1p_2cdots p_k \ 0 &other end{cases}μ(n)=⎩⎪⎨⎪⎧​1(−1)k0​n=1n=p1​p2​⋯pk​other​

p_i (i = 1, 2, cdots, k)pi​(i=1,2,⋯,k) are different prime numbers.

Given two integers mm, nn, please calculate sum_{i = 1}^{m} mu(in)∑i=1m​μ(in).

Input

One line includes two integers m (1 le m le 2e9)m(1≤m≤2e9), n (1 le n le 1e12)n(1≤n≤1e12).

Output

One line includes the answer .

样例输入复制

2 2

样例输出复制

-1

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

min_25筛的话，先得到下面这个公式

首先答案可以化简为这个

根据积性函数前缀和可以得到这个

(S_f(m,1)+1)*mu(n)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e6+5;
ll p[N],w[N],pri[N],g[N],M,m,n;
int block,id[N],sz,tot;
bool is_p[N],v[N];
void init(int n)
{
    sz=0;
    for(int i=2; i<=n; i++)
    {
        if(!is_p[i])p[++sz]=i;
        for(int j=1; j<=sz&&p[j]*i<n; j++)
        {
            is_p[i*p[j]]=1;
            if(i%p[j]==0) break;
        }
    }
}
void sieve_g(ll n)
{
    M=0;
    for(ll i=1,lst; i<=n; i=lst+1)
    {
        ll len=n/i;
        lst=n/len;
        w[++M]=len;
        g[M]=1-w[M];
        if(len<=block) id[len]=M;
    }
    for(int i=1; i<=sz; i++)
        for(int j=1; j<=M&&p[i]*p[i]<=w[j]; j++)
        {
            int op=w[j]/p[i]<=block?id[w[j]/p[i]]:n/(w[j]/p[i]);
            g[j]-=g[op]+i-1;
        }
}
ll S(ll x,ll y)
{
    ll k,ans=0;
    if(x<=1||p[y]>x) return 0;
    if(x>block) k=m/x;
    else k=id[x];
    ans=g[k]+y-1;
    for(int i=1; i<=tot&&pri[i]<=w[k]; i++)
        if(pri[i]>p[y-1]) ans++;
    for(int i=y; i<=sz&&p[i]*p[i]<=x; i++)
        if(v[i]==0) ans-=S(x/p[i],i+1);
    return ans;
}
int main()
{
    tot=0;
    cin>>m>>n;
    block=sqrt(m+0.5);
    init(m<=1e6?1e4:1e6);
    sieve_g(m);
    for(int i=1; p[i]*p[i]<=n&&i<=sz;i++)
    {
        if(n%p[i])continue;
        pri[++tot]=p[i];
        n/=p[i];
        v[i]=1;
        if(n%p[i]==0)
        {
            puts("0");
            return 0;
        }
    }
    if(n>1)
    {
        pri[++tot]=n;
        for(int i=sz; i>0; i--)
            if(p[i]==n)
            {
                v[i]=1;
                break;
            }
    }
    int t=tot&1?-1:1;
    cout<<(S(m,1)+1)*t;
    return 0;
}

莫比乌斯+杜教筛

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e6+5;
int pri[N],num;
int vst[N],mu[N],mu_n;
inline void Pre()
{
    mu[1]=1;
    for(int i=2; i<N; i++)
    {
        if(!vst[i]) pri[++num]=i,mu[i]=-1;
        for(int j=1; j<=num && i*1LL*pri[j]<N; j++)
        {
            vst[i*pri[j]]=1;
            if(i%pri[j]==0)
            {
                mu[i*pri[j]]=0;
                break;
            }
            mu[i*pri[j]]=mu[i]*mu[pri[j]];
        }
    }
    for(int i=1; i<N; i++) mu[i]+=mu[i-1];
}
unordered_map<ll,int> S;
inline int Sum(ll n)
{
    if(n<N) return mu[n];
    if(S.count(n)) return S[n];
    int tem=1;
    ll l,r;
    for(l=2; l*l<=n; l++) tem-=Sum(n/l);
    for(ll t=n/l; l<=n; l=r+1,t--)
    {
        r=n/t;
        tem-=(r-l+1)*Sum(t);
    }
    return S[n]=tem;
}
ll f(ll m,ll n)
{
    if(m==0) return 0;
    if(n==1) return Sum(m);
    int flag=1;
    for(int i=1; pri[i]*1LL*pri[i]<=n; i++)
    {
        if(n%pri[i]==0)
        {
            flag=0;
            return -f(m,n/pri[i])+f(m/pri[i],n);
        }
    }
    if(flag)return -f(m,n/n)+f(m/n,n);
}
int judge(ll n)
{
    for(int i=1; pri[i]*1LL*pri[i]<=n; i++)
    {
        if(n%pri[i]==0)
        {
            int cnt=0;
            while(n%pri[i]==0)n/=pri[i],cnt++;
            if(cnt>=2)return 0;
        }
    }
    return 1;
}
int main()
{
    Pre();
    ll m,n;
    cin>>m>>n;
    if(!judge(n))
    {
        printf("0
");
        return 0;
    }
    cout<<f(m,n);;
    return 0;
}