【学习笔记】莫比乌斯反演

其实引理用得更多，人麻了

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引理：

$$\begin{array}{rl}f(n)& =\sum _{d\mid n}\mu (d)\\ \Rightarrow f(n)& =\{\begin{array}{l}1,n=1\\ 0,else\end{array}\end{array}$$

容斥定理易证

原式：

$$\begin{array}{rl}f(n)& =\sum _{d\mid n}F(d)\\ \Rightarrow F(n)& =\sum _{d\mid n}\mu (\frac{n}{d})f(d)\end{array}$$

证明：

$$\begin{array}{rl}F(n)& =\sum _{d\mid n}\mu (\frac{n}{d})\sum _{p\mid d}F(p)\\ 令t=& \frac{d}{p}\\ & =\sum _p\sum _{t\mid \frac{n}{p}}\mu (\frac{n}{tp})F(p)\\ & =\sum _{p}F(p)\sum _{t\mid \frac{n}{p}}\mu (\frac{\frac{n}{p}}{t})\\ & =F(n)\end{array}$$

推论：

$$\begin{array}{rl}f(n)& =\sum _{n\mid d}F(d)\\ \Rightarrow F(n)& =\sum _{n\mid d}\mu (\frac{d}{n})f(d)\end{array}$$

证明：

类似于原式。

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例题

一、$YY$ 的 $GCD$

$$\begin{array}{rl}& \sum _{prime}\sum _{x\in [1,n]}\sum _{y\in [1,m]}[(x,y)==prime]\end{array}$$

令 $f(i)$ 表示 $gcd=i$ 的数的对数， $F(i)$ 表示 $i\mid gcd$ 的数的对数

$$\begin{array}{rl}F(i)& =\sum _{i\mid j}f(j)\\ f(i)& =\sum _{i\mid j}\mu (\frac{j}{i})F(j)\\ answer& =\sum _{prime}f(prime)\\ & =\sum _{prime}\sum _{prime\mid j}\mu (\frac{j}{prime})\lfloor \frac{n}{j}\rfloor \lfloor \frac{m}{j}\rfloor \\ & =\sum _j\lfloor \frac{n}{j}\rfloor \lfloor \frac{m}{j}\rfloor \sum _{prime\mid j}\mu (\frac{j}{prime})\\ 令h(x)=& \sum _{prime\mid x}\mu (\frac{x}{prime})\\ & =\sum _j\lfloor \frac{n}{j}\rfloor \lfloor \frac{m}{j}\rfloor h(j)\end{array}$$

把 $h$ 打表出来，整数分块，结束了。

//author : LH ——Who just can eat S??t
//worship WJC ——Who can f??k tourist up and down and loves 周歆恬
//worship YJX ——Who can f??k WJC up and down
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... args) {
    read (x); read (args...);
}
char For_Print[25];
template <typename T>
void write (T x) {
    if (x == 0) { putchar ('0'); return; }
    if (x < 0) { putchar ('-'); x = -x; }
    int poi = 0;
    while (x) {
        For_Print[++poi] = x % 10 + '0';
        x /= 10;
    }
    while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
    write (x); putchar (ch);
}

const LL Mod = 1e9 + 7;

LL square (LL x) { return (x * x) % Mod; }
void DEL (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void ADD (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

const int Maxn = 1e7;

int t, n, m;

int cnt, primes[Maxn + 5];
int mu[Maxn + 5], px[Maxn + 5], h[Maxn + 5];
LL pre[Maxn + 5];
bool vis[Maxn + 5];
void Euler () {
    mu[1] = 1;
    rep (i, 2, Maxn) {
        if (!vis[i]) {
            primes[++cnt] = i;
            mu[i] = -1;
            px[i] = i;
        }
        rep (j, 1, cnt) {
            if (primes[j] > Maxn / i) break;
            vis[i * primes[j]] = 1;
            if (i % primes[j] == 0) {
                mu[i * primes[j]] = 0;
                px[i * primes[j]] = primes[j];
                break;
            }
            mu[i * primes[j]] = -mu[i];
            px[i * primes[j]] = primes[j];
        }
    }
    rep (i, 1, Maxn) {
        int tmp = i;
        while (tmp != 1) {
            int prm = px[tmp];
            h[i] += mu[i / prm];
            while (tmp % prm == 0)
                tmp /= prm;
        }
    }
    rep (i, 1, Maxn) {
        pre[i] = pre[i - 1] + h[i];
    }
}

int main () {
	// freopen ("D:\\lihan\\1.in", "r", stdin);
	// freopen ("D:\\lihan\\1.out", "w", stdout);

    Euler ();

    read (t);
    while (t--) {
        read (n, m);
        if (n > m) swap (n, m);
        int l = 1, r = n;
        LL res = 0;
        while (l <= n) {
            r = Min (n / (n / l), m / (m / l));
            res += (LL)(n / l) * (m / l) * (pre[r] - pre[l - 1]);
            l = r + 1;
        }
        cout << res << endl;
    }
    return 0;
}

---

二、于神之怒

$$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j{)}^k\\ =& \sum _d\sum _i\sum _j{d}^k\sum _{p\mid \frac{gcd(i,j)}{d}}\mu (p)\\ =& \sum _p\mu (p)\sum _d{d}^k\lfloor \frac{n}{dp}\rfloor \lfloor \frac{m}{dp}\rfloor \\ =& \sum _d{d}^k\sum _p\mu (p)\lfloor \frac{n}{dp}\rfloor \lfloor \frac{m}{dp}\rfloor \\ 令& t=dp\\ =& \sum _t\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor \sum _{d\mid t}{d}^k\mu (\frac{t}{d})\\ 令& h(x)=\sum _{d\mid x}{d}^k\mu (\frac{x}{d})\\ =& \sum _t\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor h(t)\end{array}$$

//author : LH ——Who just can eat S??t
//worship WJC ——Who can f??k tourist up and down and loves 周歆恬
//worship YJX ——Who can f??k WJC up and down
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... args) {
    read (x); read (args...);
}
char For_Print[25];
template <typename T>
void write (T x) {
    if (x == 0) { putchar ('0'); return; }
    if (x < 0) { putchar ('-'); x = -x; }
    int poi = 0;
    while (x) {
        For_Print[++poi] = x % 10 + '0';
        x /= 10;
    }
    while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
    write (x); putchar (ch);
}

const LL Mod = 1e9 + 7;

LL square (LL x) { return (x * x) % Mod; }
void DEL (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void ADD (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

const int Maxn = 5 * 1e6;

int t, k, n, m;

int cnt, primes[Maxn + 5];
LL h[Maxn + 5], val[Maxn + 5], pre[Maxn + 5];
bool vis[Maxn + 5];
LL quick_pow (LL x, LL y) {
    LL res = 1;
    while (y) {
        if (y & 1) res = (res * x) % Mod;
        x = (x * x) % Mod; y >>= 1;
    }
    return res;
}
LL md (LL x, LL M) {
	return (x % M + M) % M;
}
void Euler () {
    pre[1] = h[1] = 1;
    rep (i, 2, Maxn) {
        if (!vis[i]) {
            primes[++cnt] = i;
            val[i] = quick_pow (i, k);
            h[i] = md (val[i] - 1, Mod);
        }
        rep (j, 1, cnt) {
            if (primes[j] > Maxn / i) break;
            vis[primes[j] * i] = 1;
            if (i % primes[j] == 0) {
                h[i * primes[j]] = h[i] * val[primes[j]] % Mod;
                break;
            }
            h[i * primes[j]] = h[i] * (val[primes[j]] - 1) % Mod;
        }
    }
    rep (i, 1, Maxn) pre[i] = (pre[i - 1] + h[i]) % Mod;
}

int main () {
//	freopen ("D:\\lihan\\1.in", "r", stdin);
//	freopen ("D:\\lihan\\1.out", "w", stdout);

    read (t, k); Euler ();
    while (t--) {
        read (n, m);
        if (n > m) swap (n, m);

        int l = 1, r;
        LL res = 0;
        while (l <= n) {
            r = Min (n / (n / l), m / (m / l));
            ADD (res, (LL)(n / l) * (m / l) % Mod * md (pre[r] - pre[l - 1], Mod) % Mod);
            l = r + 1;
        }
        print (res, '\n');
    }
    return 0;
}

---

三、Crash的数字表格

$$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^m\frac{i\cdot j}{gcd(i,j)}\\ =& \sum _d\sum _i\sum _j\frac{i\cdot j}{d}[gcd(i,j)==d]\\ =& \sum _d\sum _i\sum _j\frac{i\cdot j}{d}\sum _{p\mid gcd(\frac{i}{d},\frac{j}{d})}\mu (p)\\ =& \sum _p\mu (p)\sum _d\sum _{dp\mid i}\sum _{dp\mid j}\frac{i\cdot j}{d}\\ =& \sum _p\mu (p)\sum _d\frac{1}{d}\sum _{dp\mid i}i\sum _{dp\mid j}j\\ 令& lm=\lfloor \frac{m}{dp}\rfloor \\ =& \sum _p\mu (p)\sum _d\frac{1}{d}\frac{(1+lm)\ast lm}{2}\cdot dp\sum _{dp\mid i}i\\ 令& ln=\lfloor \frac{n}{dp}\rfloor \\ =& \sum _p\mu (p)\sum _d\frac{(1+lm)\ast lm}{2d}\cdot dp\cdot \frac{(1+ln)\ast ln}{2}\cdot dp\\ =& \sum _p\mu (p)\sum _d\frac{(1+lm)\ast lm\cdot (1+ln)\ast ln\cdot p^2\cdot d}{4}\\ =& \sum _p\mu (p)\sum _d\frac{(1+lm)\ast lm\cdot (1+ln)\ast ln\cdot p^2\cdot d}{4}\\ 令& t=dp\\ =& \sum _p\mu (p)\cdot p^2\frac{{\sum }_{p\mid t}t\cdot (1+\lfloor \frac{m}{t}\rfloor )\ast \lfloor \frac{m}{t}\rfloor \cdot (1+\lfloor \frac{n}{t}\rfloor )\ast \lfloor \frac{n}{t}\rfloor }{4p}\\ =& \sum _p\mu (p)\cdot p\frac{{\sum }_{p\mid t}t\cdot (1+\lfloor \frac{m}{t}\rfloor )\ast \lfloor \frac{m}{t}\rfloor \cdot (1+\lfloor \frac{n}{t}\rfloor )\ast \lfloor \frac{n}{t}\rfloor }{4}\\ =& \sum _t\frac{t\cdot (1+\lfloor \frac{m}{t}\rfloor )\ast \lfloor \frac{m}{t}\rfloor \cdot (1+\lfloor \frac{n}{t}\rfloor )\ast \lfloor \frac{n}{t}\rfloor }{4}\sum _{p\mid t}\mu (p)\cdot p\\ 令& h(t)=t\cdot \sum _{p\mid t}\mu (p)\cdot p\\ =& \sum _t\frac{(1+\lfloor \frac{m}{t}\rfloor )\ast \lfloor \frac{m}{t}\rfloor \cdot (1+\lfloor \frac{n}{t}\rfloor )\ast \lfloor \frac{n}{t}\rfloor }{4}h(t)\end{array}$$

//author : LH ——Who just can eat S??t
//worship WJC ——Who can f??k tourist up and down and loves 周歆恬
//worship YJX ——Who can f??k WJC up and down
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... args) {
    read (x); read (args...);
}
char For_Print[25];
template <typename T>
void write (T x) {
    if (x == 0) { putchar ('0'); return; }
    if (x < 0) { putchar ('-'); x = -x; }
    int poi = 0;
    while (x) {
        For_Print[++poi] = x % 10 + '0';
        x /= 10;
    }
    while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
    write (x); putchar (ch);
}

const LL Mod = 20101009;
const LL inv_2 = 10050505;

LL square (LL x) { return (x * x) % Mod; }
void DEL (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void ADD (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

const int Maxq = 1e6;
const int Maxn = 1e7;

int cnt, primes[Maxq + 5];
LL h[Maxn + 5], pre[Maxn + 5];
bool vis[Maxn + 5];
LL md (LL x, LL M) {
    return (x % M + M) % M;
}
void Euler () {
    h[1] = 1;
    rep (i, 2, Maxn) {
        if (!vis[i]) {
            primes[++cnt] = i;
            h[i] = 1 - i;
        }
        rep (j, 1, cnt) {
            if (primes[j] > Maxn / i) break;
            vis[primes[j] * i] = 1;
            if (i % primes[j] == 0) {
                h[i * primes[j]] = h[i];
                break;
            }
            h[i * primes[j]] = h[i] * (1 - primes[j]) % Mod;
        }
    }

    rep (i, 1, Maxn) pre[i] = md (pre[i - 1] + h[i] * i % Mod, Mod);
}

LL n, m;

int main () {
	// freopen ("D:\\lihan\\1.in", "r", stdin);
	// freopen ("D:\\lihan\\1.out", "w", stdout);

    Euler ();

    read (n, m);
    // cerr << n << " " << m << endl;
    if (n > m) swap (n, m);
    int l = 1, r;
    LL res = 0;
    while (l <= n) {
        LL ln = n / l, lm = m / l;
        r = Min (n / ln, m / lm);
        ADD (res, (1 + ln) * ln % Mod * (1 + lm) % Mod * lm % Mod * md (pre[r] - pre[l - 1], Mod) % Mod * inv_2 % Mod * inv_2 % Mod);
        l = r + 1;
    }
    cout << res;
    return 0;
}

---

四、$「SDOI2014」$ 数表

$$\begin{array}{rl}answer& =\sum _{i=1}^n\sum _{j=1}^m[\sigma (gcd(i,j))\le a]\sigma (gcd(i,j))\\ 令g=gc& d(i,j)\\ & =\sum _g\sum _i\sum _j[\sigma (g)\le a]\sigma (g)\\ & =\sum _g[\sigma (g)\le a]\sum _i\sum _j\sum _{d\mid gcd(\frac{i}{g},\frac{j}{g})}\mu (d)\sigma (g)\\ & =\sum _g[\sigma (g)\le a]\sum _d\mu (d)\sum _{dg\mid i}\sum _{dg\mid j}\sigma (g)\\ & =\sum _g[\sigma (g)\le a]\sum _d\mu (d)\lfloor \frac{n}{dg}\rfloor \lfloor \frac{m}{dg}\rfloor \sigma (g)\\ & =\sum _g\sum _d[\sigma (g)\le a]\cdot \mu (d)\lfloor \frac{n}{dg}\rfloor \lfloor \frac{m}{dg}\rfloor \sigma (g)\\ 令t=dg& \\ & =\sum _{g\mid t}\sum _t[\sigma (g)\le a]\cdot \mu (\frac{t}{g})\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor \sigma (g)\\ & =\sum _{g\mid t}\sum _t[\sigma (g)\le a]\cdot \mu (\frac{t}{g})\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor \sigma (g)\\ & =\sum _t\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor \sum _{g\mid t}[\sigma (g)\le a]\cdot \mu (\frac{t}{g})\sigma (g)\\ 令h(t)=& \sum _{g\mid t}[\sigma (g)\le a]\cdot \mu (\frac{t}{g})\sigma (g)\\ & =\sum _t\lfloor \frac{n}{t}\rfloor \lfloor \frac{m}{t}\rfloor h(t)\end{array}$$

我们离线求解，按照 $a$ 排序，求解完 $i-1$ 后，会加入一些 $g$，满足 $a_{i-1}\le \sigma (g)\le a_i$，我们把这些 $g$ 用类似埃筛的方法加入 $h(k\cdot g)(k\in \mathrm{Z},k\cdot g\le min(n,m))$，然后用树状数组动态维护前缀和就行了。

好妙， $DJ$ 是神仙。

//author : LH ——Who just can eat S??t
//worship WJC ——Who can f??k tourist up and down and loves 周歆恬
//worship YJX ——Who can f??k WJC up and down
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... args) {
    read (x); read (args...);
}
char For_Print[25];
template <typename T>
void write (T x) {
    if (x == 0) { putchar ('0'); return; }
    if (x < 0) { putchar ('-'); x = -x; }
    int poi = 0;
    while (x) {
        For_Print[++poi] = x % 10 + '0';
        x /= 10;
    }
    while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
    write (x); putchar (ch);
}

const LL Mod = 1ll << 31;

LL square (LL x) { return (x * x) % Mod; }
void DEL (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void ADD (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

const int Maxn = 1e5;

int t;
LL ans[Maxn + 5];

int cnt, primes[Maxn + 5];
LL px[Maxn + 5], sigma[Maxn + 5], mu[Maxn + 5];
bool vis[Maxn + 5];
deque <PII> c;
void Euler () {
    mu[1] = 1; sigma[1] = 1;
    rep (i, 2, Maxn) {
        if (!vis[i]) {
            primes[++cnt] = i;
            sigma[i] = i + 1;
            mu[i] = -1;
            px[i] = i;
        }
        rep (j, 1, cnt) {
            if (primes[j] > Maxn / i) break;
            vis[i * primes[j]] = 1;
            if (i % primes[j] == 0) {
                mu[i * primes[j]] = 0;
                px[i * primes[j]] = px[i] * primes[j];
                sigma[i * primes[j]] = sigma[i / px[i]] * (sigma[px[i]] + px[i] * primes[j]);
                break;
            }
            mu[i * primes[j]] = -mu[i];
            px[i * primes[j]] = primes[j];
            sigma[i * primes[j]] = sigma[i] * (primes[j] + 1);
        }
    }
    rep (i, 1, Maxn)
        c.push_back (MP (sigma[i], i));
    sort (c.begin (), c.end ());
}

struct qst {
    int n, m, id; LL a;
}q[Maxn + 5];
bool cmp (qst x, qst y) {
    return x.a < y.a;
}

LL BIT[Maxn + 5];
LL md (LL x, LL M) {
    return (x % M + M) % M;
}
int lowbit (int x) { return x & -x; }
void Update (int Index, LL x) {
    for (int i = Index; i <= Maxn; i += lowbit (i))
        BIT[i] += x;
}
LL Sum (int Index) {
    LL res = 0;
    for (int i = Index; i >= 1; i -= lowbit (i))
        res += BIT[i];
    return res;
}
LL Query (int l, int r) {
    return md (Sum (r) - Sum (l - 1), Mod);
}
void Add (int x) {
    for (int i = x; i <= Maxn; i += x)
        Update (i, mu[i / x] * sigma[x]);
}

int main () {
// 	freopen ("D:\\lihan\\1.in", "r", stdin);
//	freopen ("D:\\lihan\\1.out", "w", stdout);

	Euler ();
    
    read (t);
    rep (i, 1, t)
        { read (q[i].n, q[i].m, q[i].a); q[i].id = i; }
    sort (q + 1, q + 1 + t, cmp);

    rep (i, 1, t) {
        while (c.size () && c.begin () -> fi <= q[i].a) 
			{ Add (c.begin () -> se); c.pop_front (); }

        int n = q[i].n, m = q[i].m;
        if (n > m) swap (n, m);
        int l = 1, r;
        LL res = 0;
        while (l <= n) {
            r = Min (n / (n / l), m / (m / l));
            ADD (res, (n / l) * (m / l) % Mod * Query (l, r) % Mod);
            l = r + 1;
        }
        ans[q[i].id] = res;
    }
    rep (i, 1, t) print (ans[i], '\n');
    return 0;
}

---

五、约数个数和

$$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{m}d(ij)\\ =& \sum _{i=1}^n\sum _{j=1}^m\sum _{p\mid ij}1\\ =& \sum _{p=1}^{nm}\sum _{i=1}^n\sum _{j=1}^m[p\mid ij]\\ =& \sum _{p=1}^{nm}\sum _{i=1}^n\sum _{j=1}^m[\frac{p}{gcd(i,p)}\mid j]\\ =& \sum _{p=1}^{nm}\sum _{i=1}^n\lfloor \frac{m}{\frac{p}{gcd(i,p)}}\rfloor \\ =& \sum _{g=1}^n\sum _{p=1}^{nm}\sum _{i=1}^n\lfloor \frac{m\cdot g}{p}\rfloor [gcd(i,p)==g]\\ =& \sum _{g=1}^n\sum _{p=1}^{nm}\sum _{i=1}^n\lfloor \frac{m\cdot g}{p}\rfloor \sum _{d\mid gcd(\frac{i}{g},\frac{p}{g})}\mu (d)\\ =& \sum _{g=1}^n\sum _{p=1}^{\frac{nm}{g}}\sum _{i=1}^{\frac{n}{g}}\lfloor \frac{m\cdot g}{p\cdot i}\rfloor \sum _{d\mid gcd(i,p)}\mu (d)\\ =& \sum _{g=1}^n\sum _{p=1}^{\frac{nm}{g}}\sum _{i=1}^{\frac{n}{g}}\sum _{d\mid gcd(i,p)}\lfloor \frac{m\cdot g}{p\cdot i}\rfloor \mu (d)\\ =& \sum _{g=1}^n\sum _d\sum _{d\mid p}^{\frac{nm}{g}}\sum _{d\mid i}^{\frac{n}{g}}\lfloor \frac{m\cdot g}{p\cdot i}\rfloor \mu (d)\\ =& \sum _{g=1}^n\sum _d\sum _{p=1}^{\frac{nm}{gd}}\sum _{i=1}^{\frac{n}{gd}}\lfloor \frac{m\cdot g}{p\cdot i\cdot d^2}\rfloor \mu (d)\\ =& \sum _{g=1}^n\sum _d\sum _{p=1}^{\frac{nm}{gd}}\sum _{i=1}^{\frac{n}{gd}}\lfloor \frac{m\cdot g}{p\cdot i\cdot d^2}\rfloor \mu (d)\end{array}$$