洛谷P4213 Sum(杜教筛)

题目描述

给定一个正整数N(Nle2^{31}-1)N(N≤231−1)

求ans_1=sum_{i=1}^nphi(i),ans_2=sum_{i=1}^n mu(i)ans1=∑i=1nϕ(i),ans2=∑i=1nμ(i)

输入输出格式

输入格式：

一共T+1行第1行为数据组数T(T<=10) 第2~T+1行每行一个非负整数N，代表一组询问

输出格式：

一共T行，每行两个用空格分隔的数ans1,ans2

输入输出样例

输入样例#1

输出样例#1：

裸的杜教筛

sum_{i=1}^{n}varphi(i) = frac{ntimes(n+1)}{2} - sum_{d=2}^{n}sum_{i=1}^{lfloorfrac{n}{d}rfloor}varphi(i)

sum_{i=1}^{n}mu(i) = 1 - sum_{d=2}^{n}sum_{i=1}^{lfloorfrac{n}{d}rfloor}mu(i)

然后直接暴力递归计算即可

#include<cstdio>
#include<map>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define LL long long 
using namespace std;
using namespace __gnu_pbds;
const int MAXN=5000030;
int N,limit=5000000,tot=0,vis[MAXN],mu[MAXN],prime[MAXN];
LL phi[MAXN];
gp_hash_table<int,LL>Aphi,Amu;
void GetMuAndPhi()
{
    vis[1]=1;phi[1]=1;mu[1]=1;
    for(int i=1;i<=limit;i++)
    {
        if(!vis[i]) prime[++tot]=i,phi[i]=i-1,mu[i]=-1;
        for(int j=1;j<=tot&&i*prime[j]<=limit;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){mu[i*prime[j]]=0; phi[i*prime[j]]=phi[i]*prime[j]; break;}
            else {mu[i*prime[j]]=-mu[i]; phi[i*prime[j]]=phi[i]*(prime[j]-1); }
        }
    }
    for(int i=1;i<=limit;i++) mu[i]+=mu[i-1],phi[i]+=phi[i-1];
}
LL SolvePhi(LL n)
{
    if(n<=limit) return phi[n];
    if(Aphi[n]) return Aphi[n];
    LL tmp=n*(n+1)/2;
    for(int i=2,nxt;i<=n;i=nxt+1)
        nxt=min(n,n/(n/i)),
        tmp-=SolvePhi(n/i)*(LL)(nxt-i+1);
    return Aphi[n]=tmp;
}
LL SolveMu(LL n)
{
    if(n<=limit) return mu[n];
    if(Amu[n]) return Amu[n];
    LL tmp=1;
    for(int i=2,nxt;i<=n;i=nxt+1)
        nxt=min(n,n/(n/i)),
        tmp-=SolveMu(n/i)*(LL)(nxt-i+1);
    return Amu[n]=tmp;
}
int main()
{
    GetMuAndPhi();
    int QWQ;
    scanf("%d",&QWQ);
    while(QWQ--)
    {
        scanf("%lld",&N);
        printf("%lld %lldn",SolvePhi(N),SolveMu(N));
    }
    return 0;
}