bzoj3944 Sum

题目链接

problem

给出一个\(n,n < 2^{31}\)。分别求

\[\sum\limits_{i=1}^n\varphi(i),\sum\limits_{i=1}^n\mu(i)\]

solution

\(\varphi(i)\)和\(\mu(i)\)都是积性函数。

且\(\varphi(p^k)=(p-1)p^{k-1}\)，所以可以直接\(min\_25\)筛了。

因为\(\varphi(p)=p-1，p是质数\)，g函数不好处理

所以将\(\varphi(p)\)分为两个函数\(f_1(p)=p,f_2(p)=1\)。然后分别求\(g\)和\(h\)。

\(\mu\)预处理就直接是\(-h\)了。

然后\(min\_25\)筛模板就行了。

rp--

code

/*
* @author: wxyww
* @date:   2019-12-25 20:16:31
* @last modified time: 2019-12-25 21:38:39
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
using namespace std;
typedef long long ll;
const int n = 500010;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}
ll m,n,pri[n],vis[n],js,tot,g[n],h[n],sum[n];
void pre() {
    for(int i = 2;i <= 500000;++i) {
        if(!vis[i]) pri[++js] = i,sum[js] = sum[js - 1] + i;
        for(int j = 1;j <= js && 1ll * pri[j] * i <= 500000;++j) {
            vis[pri[j] * i] = 1;
            if(i % pri[j] == 0) break;
        }
    }
}
ll w[n],id1[n],id2[n];
ll sphi(ll now,ll x) {
    if(now <= 1 || pri[x] > now) return 0;
    
    ll k;

    if(now <= m) k = id1[now];
    else k = id2[n / now];
    ll ret = g[k] - h[k] - sum[x - 1] + x - 1;

    for(int k = x;k <= tot && pri[k] * pri[k] <= now;++k) {
        ll p = pri[k];
        for(int e = 1;p * pri[k] <= now;++e,p *= pri[k]) {
            ret += (pri[k] - 1) * (p / pri[k]) * sphi(now / p,k + 1) + p * (pri[k] - 1);
        }
    }
    return ret;
}
ll smu(ll now,ll x) {
    if(now <= 1 || pri[x] > now) return 0;

    ll  k;
    if(now <= m) k = id1[now];
    else k = id2[n / now];

    ll ret = -h[k] + x - 1;
    for(int k = x;k <= tot && pri[k] * pri[k] <= now;++k) {

            ret -= smu(now / pri[k],k + 1);
    } 
    return ret;
}
void solve() {
    m = sqrt(n);
    if(!n) return (void)puts("0 0");
    tot = 0;
    // memset(g,0,sizeof(g));memset(h,0,sizeof(h));

    for(ll l = 1,r;l <= n;l = r + 1) {
        r = n / (n / l);
        w[++tot] = n / l;
        if(n / l <= m) id1[n / l] = tot;
        else id2[r] = tot;

        g[tot] = ((w[tot] + 2) * (w[tot] - 1)) / 2;
        h[tot] = w[tot] - 1;

    }
    // puts("!!");
    for(int j = 1;j <= js && pri[j] <= m;++j) {
        for(int i = 1;i <= tot && 1ll * pri[j] * pri[j] <= w[i];++i) {
            ll tmp = w[i] / pri[j];
            int k;
            if(tmp <= m) k = id1[tmp];
            else k = id2[n / tmp];

            g[i] -= pri[j] * (g[k] - sum[j - 1]);
            h[i] -= h[k] - j + 1;
        }
    }

    // for(int i = 1;i <= tot;++i) printf("%lld ",h[i]);
    // puts("");
    // puts("!");
    cout<<sphi(n,1) + 1 <<" "<<smu(n,1) + 1<<endl;
    // puts("!!!");
}
int main() {
    int t = read();
    
    pre();

    while(t--) {
        n = read();solve();
    }

    return 0;
}