【JZOJ5250】【GDOI2018模拟8.11】质数

Description

Data Constraint

Solution

我们发现2f(n)就等于∑i|n[gcd(i,n/i)==1]
所以题目就是求：

∑i=1n∑j|i[gcd(j,i/j)==1]

=∑j=1n∑i=1⌊n/j⌋[gcd(i,j)==1]

=∑j=1n∑i=1⌊n/j⌋∑d|gcd(i,j)μ(d)

=∑d=1n√μ(d)∑k=1⌊n/d⌋⌊⌊n/d2⌋k⌋

枚举d，然后分块计算后面即可。

Code

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>#define ll long longusing namespace std;const ll maxn=1e6+5,mo=998244353; ll bz[maxn],u[maxn],n,i,t,j,k,l,x,y,z,d[maxn],ans,p[maxn];ll dg(ll n){    ll t,i=1,x=0;    while (i<=n){        t=n/(n/i);        x=x+n/i*(t-i+1)%mo;i=t+1;    }    return x%mo;}int main(){//  freopen("data.in","r",stdin);    scanf("%lld",&n);u[1]=1;t=sqrt(n);    for (i=2;i<=t;i++){        if (!bz[i]) bz[i]=1,d[++d[0]]=i,u[i]=-1;        for (j=1;j<=d[0];j++){            if (i*d[j]>t) break;            u[i*d[j]]=-u[i];bz[i*d[j]]=1;            if (i%d[j]==0){                u[i*d[j]]=0;                break;            }        }    }    for (i=1;i<=t;i++)        ans=ans+u[i]*dg(n/(i*i));    ans=(ans%mo+mo)%mo;    printf("%lld\n",ans);}