『题解』Luogu-P5572 [CmdOI2019]简单的数论题
思路基本同 P4240 毒瘤之神的考验。
Description
-
多测，\(T\) 组数据。
-
给定整数 \(n, m\)，请求出
\[\left[\sum_{i = 1}^n \sum_{j = 1}^m \varphi\left(\dfrac{\operatorname{lcm}(i, j)}{\gcd(i, j)} \right) \right] \bmod 23333 \] -
\(T\le 3\times 10^4, m\le n\le 5\times 10^4\)。
Solution
不妨设 \(n\le m\)。
\[\begin{aligned}
ans
& = \sum_{d = 1}^n \sum_{i = 1}^n \sum_{j = 1}^m \varphi\left(\dfrac{ij}{d^2} \right) [\gcd(i, j) = d] \\
& = \sum_{d = 1}^n \sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \sum_{j = 1}^{\left\lfloor\frac{m}{d}\right\rfloor} \varphi(ij) [\gcd(i, j) = 1] \\
& = \sum_{d = 1}^n \sum_{i = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \varphi(i) \sum_{j = 1}^{\left\lfloor\frac{m}{d}\right\rfloor} \varphi(j) [\gcd(i, j) = 1] \\
& = \sum_{d = 1}^n \sum_{k = 1}^{\left\lfloor\frac{n}{d}\right\rfloor} \mu(k) \sum_{i = 1}^{\left\lfloor\frac{n}{dk}\right\rfloor} \varphi(ik) \sum_{j = 1}^{\left\lfloor\frac{m}{dk}\right\rfloor} \varphi(jk) \\
& = \sum_{T = 1}^n \sum_{k\mid T} \mu(k) \sum_{i = 1}^{\left\lfloor\frac{n}{T}\right\rfloor} \varphi(ik) \sum_{j = 1}^{\left\lfloor\frac{m}{T}\right\rfloor} \varphi(jk)
\end{aligned}
\]
套路地设出来
\[f(k, n) = \sum_{i = 1}^n \varphi(ik)
\]
那么
\[ans = \sum_{T = 1}^n \sum_{k\mid T} \mu(k) f\left(k, \left\lfloor\dfrac{n}{T}\right\rfloor \right) f\left(k, \left\lfloor\dfrac{m}{T}\right\rfloor \right)
\]
又有递推式
\[f(k, n) = f(k, n - 1) + \varphi(nk)
\]
即 \(f\) 可以在 \(\Theta(n\ln n)\) 时间内预处理。
但是这个 \(f\) 带一个 \(k\)，无法整除分块，只能暴力，所以查询是 \(\Omicron(T\sum\limits_{i = 1}^n \dfrac{n}{i}) \simeq \Omicron(T n\ln n)\) 的。
我们想要整除分块，那就弄个前缀和。
把整个表示出来
\[g(n, m, l) = \sum_{i = 1}^l \sum_{k\mid i} \mu(k) f(k, n) f(k, m)
\]
那么整除分块大概长这样
\[g\left(\left\lfloor\dfrac{n}{l}\right\rfloor, \left\lfloor\dfrac{m}{l}\right\rfloor, r \right) - g\left(\left\lfloor\dfrac{n}{l}\right\rfloor, \left\lfloor\dfrac{m}{l}\right\rfloor, l - 1 \right)
\]
同样有递推式
\[g(n, m, l) = g(n, m, l - 1) + \sum_{k\mid l} \mu(k) f(k, n) f(k, m)
\]
预处理是
\[\Omicron\left(n \sum_{i = 1}^n \left(\sum_{j = 1}^{\frac{n}{i}} \dfrac{n}{j}\right) + \dfrac{n}{i} \right) \simeq \Omicron(n^3)
\]
会 \(\rm TLE + MLE\)。
但是它能够做到十分优秀的 \(\Omicron(T \sqrt{n})\) 回答。
如果要将两种方法结合，那么就是大家喜闻乐见的 根号分治 了。
（其实题目背景已经有提示了）
A:“数论,分块。”
设 \(k\) 为阈值，表示预处理 \(g(n, m, l)\) 时 \(n, m\) 只处理到 \(k\)。
那么计算时对于 \(l \le \left\lfloor\dfrac{n}{k}\right\rfloor\) 的直接暴力；对于 \(l > \left\lfloor\dfrac{n}{k}\right\rfloor\) 的部分，有 \(\left\lfloor\dfrac{n}{l}\right\rfloor \le k\)，直接用预处理过的 \(g\)。
Code
//18 = 9 + 9 = 18.
#include <iostream>
#include <cstdio>
#include <vector>
#define Debug(x) cout << #x << "=" << x << endl
typedef long long ll;
using namespace std;
const int MAXN = 5e4 + 5;
const int N = 5e4;
const int MAXK = 155;
const int K = 150;
const int MOD = 23333;
typedef int arr[MAXN];
int add(int a, int b) {return (a + b) % MOD;}
int sub(int a, int b) {return (a - b + MOD) % MOD;}
int mul(int a, int b) {return (ll)a * b % MOD;}
arr p, mu, phi;
bool vis[MAXN];
vector<int> f[MAXN], g[MAXK][MAXK];
void pre()
{
mu[1] = phi[1] = 1;
for (int i = 2; i <= N; i++)
{
if (!vis[i])
{
p[++p[0]] = i;
mu[i] = MOD - 1;
phi[i] = (i - 1) % MOD;
}
for (int j = 1; j <= p[0] && i * p[j] <= N; j++)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
{
mu[i * p[j]] = 0;
phi[i * p[j]] = mul(phi[i], p[j]);
break;
}
mu[i * p[j]] = mul(mu[i], mu[p[j]]);
phi[i * p[j]] = mul(phi[i], phi[p[j]]);
}
}
for (int k = 1; k <= N; k++)
{
f[k].resize(N / k + 5);
for (int n = 1; n <= N / k; n++)
{
f[k][n] = add(f[k][n - 1], phi[n * k]);
}
}
for (int n = 1; n <= K; n++)
{
for (int m = n; m <= K; m++)
{
g[n][m].resize(N / m + 5);
for (int i = 1; i <= N / m; i++)
{
int wjy = mul(mu[i], mul(f[i][n], f[i][m]));
for (int j = 1; j <= N / m / i; j++)
{
g[n][m][i * j] = add(g[n][m][i * j], wjy);
}
}
for (int l = 1; l <= N / m; l++)
{
g[n][m][l] = add(g[n][m][l], g[n][m][l - 1]);
}
}
}
}
int block(int n, int m)
{
if (n > m)
{
swap(n, m);
}
int res = 0, xsy = m / K;
for (int i = 1; i <= xsy; i++)
{
for (int j = 1; j <= xsy / i; j++)
{
int t = i * j;
res = add(res, mul(mu[i], mul(f[i][n / t], f[i][m / t])));
}
}
for (int l = xsy + 1, r; l <= n; l = r + 1)
{
int k1 = n / l, k2 = m / l;
r = min(n / k1, m / k2);
res = add(res, sub(g[k1][k2][r], g[k1][k2][l - 1]));
}
return res;
}
int main()
{
pre();
int t;
scanf("%d", &t);
while (t--)
{
int n, m;
scanf("%d%d", &n, &m);
printf("%d\n", block(n, m));
}
return 0;
}
