洛谷 P3911
P3911 最小公倍数之和
题目链接:P3911 最小公倍数之和
题目大意
简洁易懂,我就不说了
solution
我们来观察这个式子, \(lcm\)怎么操作啊? 那我们就转化成\(gcd\)
\(\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n lcm(A_i, A_j) \iff \sum\limits_{i = 1}^n\sum\limits_{j = 1}^n \dfrac{A_iA_j}{gcd(A_i, A_j)}\)
那还是很难处理啊, 那我们怎么办呢, 我们是不是可以转化成值域来做啊,记录一下每个数的个数.
我们发现这样是可行的, 毕竟 \(A_i \in [1, 50000]\)
我们设 \(N\) 为值域的最大值, \(C_i\) 是每个元素的个数, 那我们要求的就变成了:
\(\sum\limits_{i = 1}^N\sum\limits_{j = 1}^N \dfrac{i \times j \times C_i \times C_j}{gcd(A_i, A_j)}\)
这不就是莫比乌斯反演的板子题么, 那我们来推一下反演过程:
枚举公约数:
\(\sum\limits_{d = 1}^N\sum\limits_{i = 1}^N\sum\limits_{j = 1}^N \dfrac{ij}{d}[gcd(A_i, A_j) == d] \times C_i \times C_j\)
将后两重循环除以 \(d\) , \(i\) 和 \(j\) 变成 \(id\) 和 \(jd\) :
\(\sum\limits_{d = 1}^N\sum\limits_{i = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{d}\right\rfloor} ijd[gcd(A_i, A_j) == 1] \times C_{id} \times C_{jd}\)
我们伟大的莫比乌斯函数来了:
\(\sum\limits_{d = 1}^N\sum\limits_{i = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{k|gcd(i, j)} \mu(k)ijd \times C_{id} \times C_{jd}\)
我们发现 \(gcd(i,j) == 1\), 共有 \(\left\lfloor\frac{N}{d}\right\rfloor\) 个, 然后改变一下 \(k\) 的枚举方式:
\(\sum\limits_{d = 1}^N\sum\limits_{i = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{k}^{\left\lfloor\frac{N}{d}\right\rfloor} \mu(k)ijd\times C_{id} \times C_{jd}\)
交换一下枚举顺序:
\(\sum\limits_{d = 1}^N\sum\limits_{k}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{i = 1}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{d}\right\rfloor} \mu(k)ijd\times C_{id} \times C_{jd}\)
然后将最后两重循环除以 \(k\), \(id, jd\) 就变成了 \(idk, jdk\) :
\(\sum\limits_{d = 1}^N\sum\limits_{k}^{\left\lfloor\frac{N}{d}\right\rfloor}\sum\limits_{i = 1}^{\left\lfloor\frac{N}{dk}\right\rfloor}\sum\limits_{j = }^{\left\lfloor\frac{N}{dk}\right\rfloor} \mu(k)ijdk^2 \times C_{idk} \times C_{jdk}\)
我们令 \(T = dk\) :
\(\sum\limits_{T = 1}^N\sum\limits_{k | T}\sum\limits_{i = 1}^{\left\lfloor\frac{N}{T}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{T}\right\rfloor} \mu(k)k T ij\times C_{iT} \times C_{jT}\)
然后交换一下枚举顺序:
\(\sum\limits_{T = 1}^N T \sum\limits_{i = 1}^{\left\lfloor\frac{N}{T}\right\rfloor}\sum\limits_{j = 1}^{\left\lfloor\frac{N}{T}\right\rfloor} ij\times C_{iT} \times C_{jT}\sum\limits_{k | T}\mu(k)k\)
把 \(ij\) 的两维合并:
\(\sum\limits_{T = 1}^N T \sum\limits_{i = 1}^{\left\lfloor\frac{N}{T}\right\rfloor} (i\times C_{iT})^2\sum\limits_{k | T}\mu(k)k\)
最后我们预处理出 \(\mu(k)k\) 就做完了.
Code:
/**
* Author: Alieme
* Data: 2020.9.8
* Problem: P3911
* 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 tot, n, ans;
int c[MAXN], mu[MAXN], prime[MAXN], sum[MAXN];
bool vis[MAXN];
inline void init() { // 预处理mu和公式
mu[1] = 1;
for (rr int i = 2; i <= 50000; i++) { // 线性筛预处理mu
if (!vis[i]) prime[++tot] = i, mu[i] = -1;
for (rr int j = 1; j <= tot; j++) {
if (i * prime[j] > 50000) break;
vis[i * prime[j]] = 1;
mu[i * prime[j]] = -mu[i];
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
break;
}
}
}
for (rr int i = 1; i <= 50000; i++) // 预处理 mu(k)*k
for (rr int j = i; j <= 50000; j += i)
sum[j] += mu[i] * i;
}
signed main() {
init();
n = read();
for (rr int i = 1; i <= n; i++) c[read()]++;
for (rr int i = 1; i <= 50000; i++) { // 暴力求解
int s = 0;
for (rr int j = 1; j <= 50000 / i; j++) s += c[i * j] * j;
ans += i * s * s * sum[i];
}
print(ans);
}
