[P5348]密码解锁

Description

给一个长度为 \(n\) 的数组 \(a[1\dots n]\) ，知足 \(\sum_{m|x}a[x] = \mu(m)\)，求 \(a[m]\)。

\(n\le 10^{18}, m\le 10^9, \frac{n}{m}\le10^9,n\geq m\)

Solution

由另外一种形式的莫比乌斯反演：

\[ \begin{aligned} a[m] &= \sum_{m|x}\mu(\frac{x}{m})\mu(x)\\ &=\sum_{i=1}^{\frac{n}{m}}\mu(i)\mu(im)\\ &=\mu(m)\sum_{i=1}^{\lfloor\frac{n}{m}\rfloor}\mu(i)^2[\gcd(i, m) = 1]\\ \end{aligned} \]

后面那个 \(\sum\) 就是在求 \(1\dots \frac{n}{m}\) 中与 \(m\) 互质且不能写成彻底平方数的倍数的个数。

相似于 [中山市选2011]彻底平方数，能够容斥求：

令 \(N = \lfloor\frac{n}{m}\rfloor\)，

\[ \begin{aligned} a[m] &=\mu(m)\sum_{i=1}^{\frac{n}{m}}\mu(i)^2[\gcd(i, m) = 1]\\ &=\mu(m)\sum_{i=1}^{\sqrt{N}}\mu(i)\sum_{j=1}^{\lfloor\frac{N}{i^2}\rfloor}[\gcd(i^2j,m)=1]\\ &=\mu(m)\sum_{i=1}^{\sqrt{N}}\mu(i)[\gcd(i,m)=1]\sum_{j=1}^{\lfloor\frac{N}{i^2}\rfloor}[\gcd(j,m)=1]\\ &=\mu(m)\sum_{i=1}^{\sqrt{N}}\mu(i)[\gcd(i,m)=1]\sum_{j=1}^{\lfloor\frac{N}{i^2}\rfloor}\sum_{d|\gcd(j,m)}\mu(d)\\ &=\mu(m)\sum_{i=1}^{\sqrt{N}}\mu(i)[\gcd(i,m)=1]\sum_{d|m}\mu(d)\lfloor\frac{\lfloor\frac{N}{i^2}\rfloor}{d}\rfloor \end{aligned} \]

而后就能够把 \(m\) 的全部约数处理出来，暴力算（复杂度上界为 \(O(T\sqrt \frac{n}{m} \sqrt m)\)，实际后面的 \(\sqrt m\) 跑不满）。

注意\(\mu\)要筛到\(\sqrt N\)复杂度才是对的（否则多一个根号）。

code

#include <bits/stdc++.h>

typedef long long LL;
typedef unsigned long long uLL;

#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;

using namespace std;

inline void proc_status()
{
    ifstream t("/proc/self/status");
    cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}

template<class T> inline T read() 
{
    register T x(0);
    register char c;
    register int f(1);
    while (!isdigit(c = getchar())) if (c == '-') f = -1;
    while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
    return x * f;
}

template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }

const int maxN = 1e5;

bool vis[maxN + 1];
vector<int> prime;
int mu[maxN + 1];
LL n, m;

void Init()
{
    mu[1] = 1;
    for (int i = 2; i <= maxN; ++i)
    {
        if (!vis[i])
        {
            prime.push_back(i);
            mu[i] = -1;
        }
        for (int j = 0; j < SZ(prime) and prime[j] * i <= maxN; ++j)
        {
            vis[prime[j] * i] = 1;
            if (i % prime[j] == 0)
                break;
            else 
                mu[i * prime[j]] = -mu[i];
        }
    }
}

int Mu(LL M)
{
    if (M <= maxN) return mu[M];
    //GO;
    int cnt = 0;
    for (LL i = 2; i * i <= M; ++i)
        if (M % i == 0)
        {
            M /= i;
            cnt++;
            if (M % i == 0)
                return 0;
        }
    if (M != 1) cnt++;
    return (cnt & 1) ? -1 : 1;
}

vector<pair<LL, int> > p;

LL calc(LL i)
{
    LL ans(0);
    for (int l = 0; l < SZ(p); ++l)
        ans += (LL)p[l].second * (((n / m) / (i * i)) / p[l].first);
    return ans;
}

void GetP(LL m)
{
    p.clear();
    for (LL d = 1; d * d <= m; ++d)
        if (m % d == 0)
        {
            p.push_back(MP(d, Mu(d)));
            if (m / d == d) continue;
            p.push_back(MP(m / d, Mu(m / d)));
        }
}

void Solve()
{
    int T = read<int>();
    while (T--)
    {
        n = read<LL>(), m = read<LL>();

        GetP(m);

        LL ans(0);
        for (LL i = 1; i * i <= (n / m); ++i)
            if (__gcd((LL)i, m) == 1)
                ans += Mu(i) * calc(i);
        ans *= Mu(m);

        printf("%lld\n", ans);
    }
}

int main() 
{
#ifndef ONLINE_JUDGE
    freopen("xhc.in", "r", stdin);
    freopen("xhc.out", "w", stdout);
#endif
    Init();
    Solve();
    return 0;
}