[BZOJ 2820]YY的GCD

Description

神犇YY虐完数论后给傻×kAc出了一题

给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对

kAc这种傻×必然不会了，于是向你来请教……

多组输入

Input

第一行一个整数T 表述数据组数

接下来T行，每行两个正整数，表示N, M

Output

T行，每行一个整数表示第i组数据的结果

Sample Input

2
10 10
100 100

Sample Output

30
2791

HINT

T = 10000

N, M <= 10000000

题解

由之前的 [HAOI 2011]Problem b ：

令 $F(d)$ 为 $d\mid gcd(i,j)$ 的数对 $(i,j)$ 个数， $f(d)$ 为 $d=gcd(i,j)$ 的数对 $(i,j)$ 个数。

由题 $$F(k)=\sum_{d=1}^{min\left\{\left\lfloor\frac{a}{k}\right\rfloor,\left\lfloor\frac{b}{k} \right\rfloor\right\}}f(kd)$$

由莫比乌斯反演定理 $F(n)=\sum_{n\mid d} f(d)\Rightarrow f(n)=\sum_{n\mid d} \mu(\frac{d}{n})F(d)$
\begin{aligned}\Rightarrow f(k)&=\sum_{d=1}^{min\left\{\left\lfloor\frac{a}{k}\right\rfloor,\left\lfloor\frac{b}{k}\right\rfloor\right\}}\mu(d)F(kd)\\&=\sum_{d=1}^{min\left\{\left\lfloor\frac{a}{k} \right\rfloor,\left\lfloor\frac{b}{k}\right\rfloor\right\}}\mu(d)\cdot\left\lfloor\frac{a}{kd}\right\rfloor\cdot\left\lfloor\frac{b}{kd}\right\rfloor\end{aligned}

显然对于此题 \begin{aligned}ans&=\sum_{isprime(p)}^{min\{a,b\}}\sum_{d=1}^{min\left\{\left\lfloor\frac{a}{p}\right\rfloor,\left\lfloor\frac{b}{p}\right\rfloor\right\}}\mu(d)\cdot\left\lfloor\frac{a}{pd}\right\rfloor\cdot\left\lfloor\frac{b}{pd}\right\rfloor\\&=\sum_{d=1}^{min\{a,b\}}\sum_{isprime(p)}^{min\left\{\left\lfloor\frac{a}{d}\right\rfloor,\left\lfloor\frac{b}{d}\right\rfloor\right\}}\mu(d)\cdot\left\lfloor\frac{a}{pd}\right\rfloor\cdot\left\lfloor\frac{b}{pd}\right\rfloor\end{aligned}

我们试着枚举 $pd$ $$ans=\sum_{pd=1}^{min\{a,b\}}\left\lfloor\frac{a}{pd}\right\rfloor\cdot\left\lfloor\frac{b}{pd}\right\rfloor\cdot\sum_{isprime(p')\wedge p'\mid(pd)}\mu\left(\frac{pd}{p'}\right)$$

我们令 $F(pd)=\sum_{isprime(p')\wedge p'\mid(pd)}\mu\left(\frac{pd}{p'}\right)$  $$ans=\sum_{pd=1}^{min\{a,b\}}F(pd)\cdot\left\lfloor\frac{a}{pd}\right\rfloor\cdot\left\lfloor\frac{b}{pd}\right\rfloor$$

只要我们预处理出 $F$ 的前缀就可以用数论分块来做了。

预处理 $F$ 我们可以线性筛后枚举质数求（详见代码）。

//It is made by Awson on 2018.1.22
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 10000000;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(LL x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

int F[N+5], a, b;
int isprime[N+5], prime[N+5], tot = 0, mu[N+5];

void get_F() {
    memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, mu[1] = 1;
    for (int i = 2; i <= N; i++) {
    if (isprime[i]) prime[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i*prime[j] <= N; j++) {
        isprime[i*prime[j]] = 0;
        if (i%prime[j]) mu[i*prime[j]] = -mu[i];
        else {mu[i*prime[j]] = 0; break; }
    }
    }
    for (int i = 1; i <= tot; i++) for (int j = 1, p = prime[i]; j*p <= N; j++) F[j*p] += mu[j];
    for (int i = 1; i <= N; i++) F[i] += F[i-1];
}
LL cal(int a, int b) {
    if (a > b) Swap(a, b); LL ans = 0;
    for (int i = 1, last; i <= a; i = last+1) {
    last = Min(a/(a/i), b/(b/i));
    ans += (LL)(F[last]-F[i-1])*(a/i)*(b/i);
    }
    return ans;
}
void work() {
    read(a), read(b); writeln(cal(a, b));
}
int main() {
    int t; read(t); get_F();
    while (t--) work();
    return 0;
}

当然我们也可以利用函数 $F$ 的积性。线性筛的过程就把 $F$ 筛出来。设枚举的数为 $i$ ，质数为 $p$ ：

1. $p\mid i$ ， $F(ip)=\mu(i)$ ；

2. $p\nmid i$ ， $F(ip)=\mu(i)-F(i)$ 。

//It is made by Awson on 2018.1.23
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 10000000;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(LL x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

int n, m; LL F[N+5];
int isprime[N+5], prime[N+5], tot, mu[N+5];

void get_F() {
    memset(isprime, 1, sizeof(isprime)); isprime[1] = 0; mu[1] = 1;
    for (int i = 2; i <= N; i++) {
    if (isprime[i]) prime[++tot] = i, mu[i] = -1, F[i] = 1;
    for (int j = 1; j <= tot && i*prime[j] <= N; j++) {
        isprime[i*prime[j]] = 0;
        if (i%prime[j]) mu[i*prime[j]] = -mu[i], F[i*prime[j]] = mu[i]-F[i];
        else {mu[i*prime[j]] = 0, F[i*prime[j]] = mu[i]; break; }
    }
    F[i] += F[i-1];
    }
}
LL cal(int n, int m) {
    if (n > m) Swap(n, m); LL ans = 0;
    for (int i = 1, last; i <= n; i = last+1) {
    last = Min(n/(n/i), m/(m/i));
    ans += (LL)(F[last]-F[i-1])*(n/i)*(m/i);
    }
    return ans;
}
void work() {
    read(n), read(m), writeln(cal(n, m));
}
int main() {
    int t; read(t); get_F();
    while (t--) work();
    return 0;
}