洛谷 P3327

题目链接:P3327 [SDOI2015]约数个数和

题目大意

求约数个数和

solution

我们知道有这么一个公式: (d(nm) = sumlimits_{i | n}sumlimits_{j | m}[gcd(i, j) == 1])

证明: 对于 (nm) 的每个质因数 (p), 设 (n = n_1 * p^a, m = m_1 * p^b), 那么 (nm = n_1 * m_1 * p^{a + b}), p 对 d(nm) 的贡献是 (a+b+1).在等式的右边里,((i_1 * p^a, j_1), (i _ 1 * p^{a - 1}, j_1), ..., (i_1, j_1), ..., (i_1, j_1 * p^b)) 都是可行的数对, 共 (a + b + 1) 个, 证毕.

那么:

(sumlimits_{i = 1}^Nsumlimits_{j = 1}^M d(i, j))

标准的展开:

$= sumlimits_{i = 1}^N sumlimits_{j = 1}^{M} sumlimits_{k | i} sumlimits_{l | j} [gcd(k, l) == 1] $

$=sumlimits_{i = 1}^N sumlimits_{j = 1}^Msumlimits_{d = 1}^{min(N, M)}mu(d)sumlimits_{k|i}sumlimits_{l|j}[gcd(k, l)|d] $

交换枚举顺序:

(=sumlimits_{d = 1}^{min(N, M)}mu(d)sumlimits_{k|i}sumlimits_{l|j}[gcd(k, l)|d] leftlfloorfrac{N}{k} ight floorleftlfloorfrac{M}{l} ight floor)

$=sumlimits_{d = 1}^{min(N, M)}mu(d)sumlimits_{k}^{leftlfloorfrac{N}{k} ight floor}sumlimits_{l}^{leftlfloorfrac{M}{l} ight floor}leftlfloorfrac{N}{dk} ight floorleftlfloorfrac{M}{dl} ight floor $

交换枚举顺序:

$=sumlimits_{d = 1}^{min(N, M)}mu(d)sumlimits_{k}^{leftlfloorfrac{N}{k} ight floor}leftlfloorfrac{N}{dk} ight floorsumlimits_{l}^{leftlfloorfrac{M}{l} ight floor}leftlfloorfrac{M}{dl} ight floor $

记 (f(i) = sumlimits_{j = 1}^i leftlfloorfrac{i}{j} ight floor), 则我们要求 $=sumlimits_{i = 1}^{min(N, M)}mu(i) * f(leftlfloorfrac{N}{i} ight floor) * f(leftlfloorfrac{M}{i} ight floor) $

然后用一下数论分块我们这道题就做完了

Code:

/**
*    Author: Alieme
*    Data: 
*    Problem: 
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define int long long
#define rr register

#define inf 1e9
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int T, n, m, ans, tot;

int mu[MAXN], d[MAXN], c[MAXN], prime[MAXN];

bool vis[MAXN];

inline void init() {
	mu[1] = d[1] = c[1] = 1;
	for (rr int i = 2; i <= 50000; i++) {
		if (!vis[i]) prime[++tot] = i, mu[i] = -1, c[i] = 1, d[i] = 2;
		for (int j = 1; j <= tot; j++) {
			if (i * prime[j] > 50000) break;
			vis[i * prime[j]] =  1;
			mu[i * prime[j]] = -mu[i];
			d[i * prime[j]] = d[i] * d[prime[j]];
			c[i * prime[j]] = 1;
			if (i % prime[j] == 0) {
				mu[i * prime[j]] = 0;
				d[i * prime[j]] = d[i] / (c[i] + 1) * (c[i] + 2);
				c[i * prime[j]] = c[i] + 1;
				break;
			}
		}
	}
	for (rr int i = 1; i <= 50000; i++) mu[i] += mu[i - 1];
	for (rr int i = 1; i <= 50000; i++) d[i] += d[i - 1];		
}

signed main() {
	init();
	T = read();
	while (T--) {
		n = read();
		m = read();
		ans = 0;
		for (rr int l = 1, r; l <= min(n, m); l = r + 1) {
			r = min(n / (n / l), m / (m / l));
			ans += (mu[r] - mu[l - 1]) * d[n / l] * d[m / l];
		}
		print(ans);
		puts("");
	}
}