luoguP4213 [模板]杜教筛

https://www.luogu.org/problemnew/show/P4213

同 bzoj3944

考虑用杜教筛求出莫比乌斯函数前缀和，第二问随便过，第一问用莫比乌斯反演来做，中间的整除分块里的莫比乌斯前缀和刚好用第二问来做

杜教筛的时候先线性筛出前 N 个数的莫比乌斯函数前缀和，其余的用 map 记忆化搜索，实测 N 取 3670000 最佳(其实我只测了3次)

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;

typedef long long ll;

template <typename _T>

inline void read(_T &f) {

    f = 0; _T fu = 1; char c = getchar();

    while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}

    while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}

    f *= fu;

}

const int N = 3670000;

map <ll, ll> val;

int mu[N], pri[N], isp[N], s[N], len = 0;

int T, n;

void Mu() {

    mu[1] = 1;

    for(int i = 2; i <= N - 10; i++) {

        if(!isp[i]) {pri[++len] = i; mu[i] = -1;}

        for(int j = 1; j <= len && i * pri[j] <= (N - 10); j++) {

            isp[i * pri[j]] = 1;

            if(i % pri[j] == 0) break;

            mu[i * pri[j]] = -mu[i];

        }

    }

    for(int i = 1; i <= N - 10; i++) s[i] = s[i - 1] + mu[i];

}

ll getmu(ll n) {

    if(n <= N - 10) return (ll)s[n];

    if(val.count(n)) return val[n];

    ll ans = 0;

    for(ll l = 2, r; l <= n; l = r + 1) {

        r = n / (n / l);

        ans += (r - l + 1) * getmu(n / l);

    }

    val[n] = 1 - ans;

    return val[n];

}

ll getphi(ll n) {

    ll ans = 0;

    for(ll l = 1, r; l <= n; l = r + 1) {

        r = n / (n / l);

        ans += (getmu(r) - getmu(l - 1)) * (n / l) * (n / l);

    }

    return (ans + 1) >> 1;

}

int main() {

    read(T); Mu();

    while(T--) {

        read(n);

        printf("%lld %lld\n", getphi(n), getmu(n));

    }

    return 0;

}