P1447 [NOI2010]能量采集
題目連結:P1447 [NOI2010]能量采集
題目大意
求 \(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j) \times 2 - 1\)
solution
如果有錯誤,歡迎指出,如不會莫比烏斯反演和狄利克雷卷積,請自行百度
\(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j) \times 2 - 1 \Rightarrow \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j) \times 2 - nm\)
那我們隻需要處理 \(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j)\) 即可,那我們怎麼辦呢?
莫比烏斯反演,這就是莫比烏斯反演的闆子題
\(f(d) = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}[\gcd(i,j) == k]\)
\(\Rightarrow f(k) = \sum\limits_{i = 1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{m}{k}\right\rfloor}[\gcd(i,j) == 1]\)
\(\Rightarrow f(k) = \sum\limits_{i = 1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{m}{k}\right\rfloor}\sum\limits_{d|\gcd(i,j)}\mu(d)\)
\(\Rightarrow f(d) = \sum\limits_{i = 1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{m}{k}\right\rfloor}\sum\limits_{d|i}\sum\limits_{d|j}\mu(d)\)
\(\Rightarrow f(k) = \sum\limits_{d = 1}^{\left\lfloor\frac{min(n, m)}{k}\right\rfloor}\mu(d)\sum\limits_{i = 1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum\limits_{d|i}\sum\limits_{j = 1}^{\left\lfloor\frac{m}{k}\right\rfloor}\sum\limits_{d|j}1\)
\(\Rightarrow f(k) = \sum\limits_{d = 1}^{\left\lfloor\frac{min(n, m)}{k}\right\rfloor}\mu(d)\left\lfloor\frac{n}{kd}\right\rfloor\left\lfloor\frac{m}{kd}\right\rfloor\)
然後我們用一下整數分塊,求出每一個 \(f(k) \times k\) 加起來就是 \(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j)\)
然後我們逾時了, 那我們怎麼辦呢?
狄利克雷卷積,能幫我們完美的解決這個問題
我們繼續操作我們推出的式子
\(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j) = \sum\limits_{k = 1}^{min(n, m)}f(k) \times k\)
\(\Rightarrow \sum\limits_{k = 1}^{min(n, m)}k \sum\limits_{d = 1}^{\left\lfloor\frac{min(n, m)}{k}\right\rfloor}\mu(d)\left\lfloor\frac{n}{kd}\right\rfloor\left\lfloor\frac{m}{kd}\right\rfloor\)
\(\Rightarrow \sum\limits_{k = 1}^{min(n, m)}k \sum\limits_{dk = 1}^{min(n, m)}\mu(d)\left\lfloor\frac{n}{kd}\right\rfloor\left\lfloor\frac{m}{kd}\right\rfloor\)
令 \(T=dk\)
\(\Rightarrow \sum\limits_{k = 1}^{min(n, m)}k \sum\limits_{T = 1}^{min(n, m)}[k|T]\mu(\frac{T}{k})\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\)
\(\Rightarrow \sum\limits_{k = 1}^{min(n, m)}\sum\limits_{k|T}^{min(n, m)}k\mu(\frac{T}{k})\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\)
\(\Rightarrow \sum\limits_{T = 1}^{min(n, m)}\sum\limits_{k|T}^{min(n, m)}k\mu(\frac{T}{k})\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\)
\(\Rightarrow \sum\limits_{T = 1}^{min(n, m)}\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\sum\limits_{k|T}^{min(n, m)}k\mu(\frac{T}{k})\)
這時候就要用上我們的狄利克雷卷積了
設 \(h(T) = \sum\limits_{d|T} d\times \mu(\frac{T}{d})\)
\(\Rightarrow h = id * \mu\)
\(\Rightarrow h * 1 = id * (\mu * 1)\)
\(\mu * 1 = \epsilon\)
\(\Rightarrow h * 1 = id * \epsilon\)
\(\Rightarrow h * 1 = id\)
\(\phi * 1 = id\)
\(\Rightarrow h = \phi\)
那麼我們帶入原式
\(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}gcd(i, j) = \sum\limits_{T = 1}^{min(n, m)} \left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor \phi(T)\)
然後我們有線性篩篩歐拉函數就好了, \(O(min(n, m))\)
Code:
/**
* Author: Alieme
* Data: 2020.9.7
* Problem: Luogu P1447
* Time: O(min(n, m))
*/
#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 n, m, ans, tot;
int phi[MAXN], sum[MAXN], prime[MAXN];
bool vis[MAXN];
inline void init() {
phi[1] = 1;
for (rr int i = 2; i <= 100000; i++) {
if (!vis[i]) prime[++tot] = i, phi[i] = i - 1;
for (rr int j = 1; j <= tot; j++) {
if (i * prime[j] > 100000) break;
vis[i * prime[j]] = 1;
phi[i * prime[j]] = phi[i] * phi[prime[j]];
if ((i % prime[j]) == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
}
}
for (rr int i = 1; i <= 100000; i++) sum[i] = sum[i - 1] + phi[i];
}
signed main() {
n = read();
m = read();
init();
for (rr int l = 1, r; l <= min(n, m); l = r + 1) {
r = min(n / (n / l), m / (m / l));
ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
}
ans = 2 * ans - n * m;
print(ans);
}
時間會刺破青春表面的彩飾,會在美人的額上掘深溝淺槽;會吃掉稀世之珍!天生麗質,什麼都逃不過他那橫掃的鐮刀。
部落客寫的那麼好,就不打賞一下麼(打賞在右邊)