Project Euler 379

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Project Euler 379

题目

Least common multiple count

Let \(f(n)\) be the number of couples \((x,y)\) with \(x\) and \(y\) positive integers, \(x \le y\) and the least common multiple of \(x\) and \(y\) equal to \(n\).

Let \(g\) be the summatory function of \(f\), i.e.: \(g(n) = \sum f(i)\) for \(1 \le i \le n\).

You are given that \(g(10^6) = 37429395\).

Find \(g(10^{12})\).

解决方案

令\(N=10^{12}\)。把 \(n\) 分解为素因子：\(\displaystyle{n=\prod_{i=1}^t p_i^{e_i}.}\)那么写\(\displaystyle{x=\prod_{i=1}^t p_i^{a_i}, y=\prod_{i=1}^t p_i^{b_i}, a_i,b_i\ge 0.}\)则\(\displaystyle{\mathrm{lcm}(x,y)=\prod_{i=1}^t p_i^{\max(a_i,b_i)}.}\) 因此 \(\mathrm{lcm}(x,y)=n\) 当且仅当对每个素数 \(p_i\) 都有\(\max(a_i,b_i)=e_i.\)

固定某个 \(p^{e}\)。满足 \(\max(a,b)=e\) 的有序对 \((a,b)\) 个数为：

  • 总共有 \((e+1)^2\) 个 \((a,b)\in[0,e]^2\)；
  • 去掉 \(\max(a,b)\le e-1\) 的，即 \([0,e-1]^2\) 共 \(e^2\) 个。

所以该素数贡献为 \((e+1)^2-e^2=2e+1\)。不同素数独立相乘，得到有序 \((x,y)\) 个数：\(\displaystyle{F(n)=\prod_{i=1}^t (2e_i+1).}\)

但注意\(n^2\)的因子个数恰好是\(\displaystyle{\tau(n^2)=\prod_{i=1}^t (2e_i+1),}\)因此\(F(n)=\tau(n^2).\)

若 \((x,y)\) 是解，则 \((y,x)\) 也是解；除非 \(x=y\)。唯一的对角线解是 \((n,n)\)。因此把有序对对半并修正对角线：\(f(n)=\dfrac{\tau(n^2)+1}{2}.\)

由上式\(\displaystyle{g(n)=\sum_{k=1}^n f(k)=\frac{1}{2}\sum_{k=1}^n \bigl(\tau(k^2)+1\bigr)=\frac{n+\sum_{k=1}^n \tau(k^2)}{2}}\).记\(\displaystyle{S(n)=\sum_{k=1}^n \tau(k^2),}\)则 \(g(n)=\dfrac{n+S(n)}{2}.\)接下来核心就是高效计算 \(S(N)\)。

令 \(\mu\) 为 Möbius 函数，\(\mu^2(n)\) 是“无平方因子（squarefree）指示函数”（若 \(n\) 是无平方因子数则为 \(1\)，否则为 \(0\)）。

那么有如下狄利克雷卷积：

\[ \tau(n^2)=(\tau \ast \mu^2)(n)=\sum_{d\mid n}\tau(d)\mu^2\left(\frac{n}{d}\right). \]

证明很直接：\(\mu^2(n)\)和\(\tau(d)\)都是乘法函数，因此只需验证素数幂 \(n=p^a\)的情况：\((\tau*\mu^2)(p^a)=\tau(p^a)\mu^2(1)+\tau(p^{a-1})\mu^2(p)=(a+1)\cdot 1+a\cdot 1=2a+1=\tau(p^{2a})=\tau((p^a)^2).\)最终得证。

由上述恒等式可得：\(\displaystyle{S(n)=\sum_{m\le n}\tau(m^2)=\sum_{m\le n}\sum_{d\mid m}\tau(d)\mu^2\left(\frac{m}{d}\right).}\)令 \(m=d\cdot k\)，交换求和次序，有\(\displaystyle{S(n)=\sum_{k\le n}\mu^2(k)\sum_{d\le n/k}\tau(d).}\)

定义“约数函数的前缀和”：\(\displaystyle{D(x)=\sum_{i\le x}\tau(i).}\)于是得到非常关键的形式：

\[ S(n)=\sum_{k\le n}\mu^2(k)D\left(\left\lfloor\frac{n}{k}\right\rfloor\right). \]

这启发我们使用数论分块解决这道题目。当 \(k\) 变化时，\(\left\lfloor n/k\right\rfloor\) 只会取 \(O(\sqrt n)\) 种不同值。令\(q=\left\lfloor\dfrac{n}{k}\right\rfloor, r=\left\lfloor\dfrac{n}{q}\right\rfloor,\)则对所有 \(k\in[k_0,r]\) 都有相同的商 \(q\)。因此\(\displaystyle{\sum_{k=k_0}^{r}\mu^2(k)=Q(r)-Q(k_0-1),}\)

其中\(\displaystyle{Q(x)=\sum_{i\le x}\mu^2(i)}\)是squarefree 计数函数。

于是我们把\(S(n)\)分成一个个块，其中每个 block 对应一段连续的 \(k\) 使得 \(\lfloor n/k\rfloor\) 相同。

为计算\(D(x)\)，有\(\displaystyle{D(x)=\sum_{i\le x}\tau(i)=\sum_{a=1}^{x}\left\lfloor\frac{x}{a}\right\rfloor=2\sum_{a=1}^{\lfloor\sqrt x\rfloor}\left\lfloor\frac{x}{a}\right\rfloor-\lfloor\sqrt x\rfloor^2.}\)

这样也可以使用数论分块完成完成。

为计算\(Q(x)\)，利用容斥原理，可得无平方因子数的个数满足\(\displaystyle{Q(x)=\sum_{d=1}^{\lfloor\sqrt x\rfloor} \mu(d)\left\lfloor\frac{x}{d^2}\right\rfloor.}\) 为了让多次查询更快，可以把 \(d\) 在立方根处截断：令 \(y=\lfloor x^{1/3}\rfloor\)。

  • 对 \(d\le y\)：直接累加 \(\mu(d)\left\lfloor x/d^2\right\rfloor\)；
  • 对 \(d>y\)：此时 \(v=\left\lfloor x/d^2\right\rfloor \le \left\lfloor x/(y+1)^2\right\rfloor = O(x^{1/3})\)。把相同的 \(v\) 归并：\(\left\lfloor\dfrac{x}{d^2}\right\rfloor=v\iff d\in\left(\sqrt{\dfrac{x}{v+1}},\sqrt{\dfrac{x}{v}}\right].\)若预先有 Mertens 前缀\(\displaystyle{M(t)=\sum_{i\le t}\mu(i),}\)则该段内 \(\sum \mu(d)=M(d_{\max})-M(d_{\min}-1)\)，从而这一部分在 \(O(x^{1/3})\) 内算完。

代码

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#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

// -------------------- globals (pure-function style) --------------------
ll G_N = 0;
int G_LIMIT = 0;

vector<int8_t> G_mu;      // mu[i]
vector<uint32_t> G_Qpref; // sum_{k<=i} mu(k)^2
vector<ll> G_Dpref;       // sum_{k<=i} tau(k)
vector<int> G_Mertens;    // M(t)=sum_{k<=t} mu(k), t<=sqrt(N)

unordered_map<ll, ll> G_cacheD;
unordered_map<ll, ll> G_cacheQ;

// -------------------- integer roots --------------------
inline ll isqrt_ll(ll x)
{
    long double r = sqrt((long double)x);
    ll y = (ll)r;
    while ((__int128)(y + 1) * (y + 1) <= x)
        ++y;
    while ((__int128)y * y > x)
        --y;
    return y;
}

inline ll icbrt_ll(ll x)
{
    long double r = cbrt((long double)x);
    ll y = (ll)llround(r);
    auto cube = [](ll t) -> __int128
    { return (__int128)t * t * t; };
    while (cube(y + 1) <= x)
        ++y;
    while (cube(y) > x)
        --y;
    return y;
}

// -------------------- sieve: mu, tau, prefixes --------------------
void build_sieve(int limit)
{
    // Linear sieve
    vector<uint32_t> lp(limit + 1, 0);  // least prime factor
    vector<uint16_t> tau(limit + 1, 0); // divisor count
    vector<uint8_t> exp(limit + 1, 0);  // exponent of lp in i
    vector<int> primes;
    primes.reserve(limit / 10);

    G_mu.assign(limit + 1, 0);
    G_mu[1] = 1;
    tau[1] = 1;

    for (int i = 2; i <= limit; ++i)
    {
        if (lp[i] == 0)
        {
            lp[i] = (uint32_t)i;
            primes.push_back(i);
            G_mu[i] = -1;
            tau[i] = 2;
            exp[i] = 1;
        }

        uint32_t li = lp[i];
        uint8_t ei = exp[i];
        uint16_t ti = tau[i];
        int8_t mui = G_mu[i];

        for (int p : primes)
        {
            long long x = 1LL * i * p;
            if (x > limit)
                break;

            lp[(int)x] = (uint32_t)p;

            if ((uint32_t)p == li)
            {
                // p divides i already
                G_mu[(int)x] = 0;
                uint8_t e = (uint8_t)(ei + 1);
                exp[(int)x] = e;
                // tau(i*p) = tau(i) / (ei+1) * (e+1)
                tau[(int)x] = (uint16_t)((ti / (ei + 1)) * (e + 1));
                break;
            }
            else
            {
                G_mu[(int)x] = (int8_t)(-mui);
                exp[(int)x] = 1;
                tau[(int)x] = (uint16_t)(ti * 2);
            }
        }
    }

    // Prefixes
    G_Qpref.assign(limit + 1, 0);
    G_Dpref.assign(limit + 1, 0);

    uint32_t sQ = 0;
    ll sD = 0;
    for (int i = 1; i <= limit; ++i)
    {
        if (G_mu[i] != 0)
            ++sQ; // mu^2(i)=1 iff mu(i)!=0
        sD += tau[i];
        G_Qpref[i] = sQ;
        G_Dpref[i] = sD;
    }
}

// -------------------- Mertens prefix up to sqrt(N) --------------------
void build_mertens_prefix(int maxT)
{
    G_Mertens.assign(maxT + 1, 0);
    int s = 0;
    for (int i = 1; i <= maxT; ++i)
    {
        s += (int)G_mu[i];
        G_Mertens[i] = s;
    }
}

// -------------------- D(x) = sum_{i<=x} tau(i) = sum_{k<=x} floor(x/k) --------------------
inline ll divisorSummatory(ll x)
{
    if (x <= G_LIMIT)
        return (ll)G_Dpref[(int)x];

    auto it = G_cacheD.find(x);
    if (it != G_cacheD.end())
        return it->second;

    // quotient grouping: O(#distinct floor(x/k)) ~ O(sqrt(x))
    ll ans = 0;
    ll l = 1;
    while (l <= x)
    {
        ll v = x / l;
        ll r = x / v;
        ans += v * (r - l + 1);
        l = r + 1;
    }

    G_cacheD.emplace(x, ans);
    return ans;
}

// -------------------- Q(x) = sum_{i<=x} mu(i)^2 = #squarefree <= x --------------------
inline ll squarefreeCount(ll x)
{
    if (x <= G_LIMIT)
        return (ll)G_Qpref[(int)x];

    auto it = G_cacheQ.find(x);
    if (it != G_cacheQ.end())
        return it->second;

    // Q(x)=sum_{d<=sqrt(x)} mu(d)*floor(x/d^2)
    // cube-root split + Mertens block for d>cuberoot(x)
    ll y = icbrt_ll(x);

    ll sum = 0;
    for (ll d = 1; d <= y; ++d)
    {
        sum += (ll)G_mu[(int)d] * (x / (d * d));
    }

    ll yy = y + 1;
    ll vmax = x / (yy * yy);
    for (ll v = 1; v <= vmax; ++v)
    {
        ll hi = isqrt_ll(x / v);
        if (hi <= y)
            break;

        ll lo = isqrt_ll(x / (v + 1)) + 1;
        if (lo <= y)
            lo = y + 1;

        // sum_{d in [lo..hi]} mu(d) = M(hi)-M(lo-1)
        sum += v * ((ll)G_Mertens[(int)hi] - (ll)G_Mertens[(int)(lo - 1)]);
    }

    G_cacheQ.emplace(x, sum);
    return sum;
}

// -------------------- main solver: g(n) = (n + S(n))/2 --------------------
ll solve_g(ll n, int limit = 15'000'000)
{
    G_N = n;
    G_LIMIT = limit;

    build_sieve(G_LIMIT);

    ll sqrt_n = isqrt_ll(G_N);
    build_mertens_prefix((int)sqrt_n);

    G_cacheD.clear();
    G_cacheQ.clear();
    G_cacheD.reserve(1 << 20);
    G_cacheQ.reserve(1 << 20);

    // S(n)=sum_{k<=n} mu(k)^2 * D(floor(n/k)) via quotient blocks
    ll S = 0;
    ll k = 1;
    ll prevQ = 0; // Q(k-1)

    while (k <= G_N)
    {
        ll q = G_N / k;
        ll k2 = G_N / q;

        ll Qk2 = squarefreeCount(k2);
        ll cnt_sqfree = Qk2 - prevQ;

        S += cnt_sqfree * divisorSummatory(q);

        prevQ = Qk2;
        k = k2 + 1;
    }

    return (G_N + S) / 2;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cout << solve_g(1'000'000'000'000LL) << "\n";
    return 0;
}