【JZOJ5250】【GDOI2018模拟8.11】质数

Description

Data Constraint

Solution

我们发现$2^{f(n)}$就等于$\sum_{i|n}[gcd(i,n/i)==1]$
所以题目就是求：

$$\sum_{i=1}^{n}\sum_{j|i}[gcd(j,i/j)==1]$$ $$=\sum_{j=1}^{n}\sum_{i=1}^{\lfloor n/j\rfloor}[gcd(i,j)==1]$$
$$=\sum_{j=1}^{n}\sum_{i=1}^{\lfloor n/j\rfloor}\sum_{d|gcd(i,j)}\mu(d)$$
$$=\sum_{d=1}^{\sqrt n}\mu(d)\sum_{k=1}^{\lfloor n/d\rfloor}{ \lfloor{\lfloor n/d^2\rfloor\over k}\rfloor}$$
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 long
using 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);
}