hdu6715 算术 2019百度之星初赛3-1003

题目地址

http://acm.hdu.edu.cn/showproblem.php?pid=6715

题解

还是不会这题的容斥做法qwq。hjw当场写了个容斥A了。我推了个莫反，但是没反应过来我的式子能\(n\log n\)暴力算...

\[\begin{aligned} &\sum_i \sum_j \mu(\frac{ij}{(i,j)})\\ &=\sum_{d} \sum_i \sum_j \mu(\frac{i}{d}) \mu(\frac{j}{d}) \mu(d) [(i,j)=d]\\ &=\sum_{d}\mu(d)\sum_i^{\frac{n}{d}} \sum_j ^\frac{m}{d} \mu(id)\mu(jd)[(i,j)=1]\\ &=\sum_{d}\mu(d)\sum_i^{\frac{n}{d}} \sum_j ^\frac{m}{d} \mu(id)\mu(jd)\sum_{k|(i,j)}\mu(k)\\ &=\sum_{d}\mu(d)\sum_{k=1}^{\frac{n}{d}} \mu(k)\sum_i^{\frac{n}{kd}} \sum_j ^\frac{m}{kd} \mu(kdi)\mu(kdj)\\ &设T=kd\\ &=\sum_T \left( \sum_{i} ^ {\frac{n}{T}} \mu(iT) \right) \left( \sum_{j} ^ {\frac{m}{T}} \mu(jT) \right)\sum_{d|T} \mu(d)\mu(\frac{T}{d}) \end{aligned} \]

第一步就是利用了\(\mu\)是个积性函数的性质，\(i\)和\(j\)除掉\((i,j)\)后显然互质，然后再乘上\((i,j)\)即可得到\(\mu(\frac{ij}{(i,j)})\)了。
然后第二步是乘上了\(\mu^2 (d)\)（当\(d\)无平方因子时，\(\mu^2 (d)=1\)，当有平方因子时本身这一项也是\(0\)），所以可以直接乘上\(\mu^2 (d)\)而不会对式子造成影响。
最后式子三个东西全都能\(n \log n\)埃筛筛出来...总复杂度\(O(T n \log n)\)
开了long long所以可能跑的比较慢...看起来是不用开的

#include <bits/stdc++.h>
using namespace std;

namespace io {
char buf[1<<21], *p1 = buf, *p2 = buf, buf1[1<<21];
inline char gc() {
    if(p1 != p2) return *p1++;
    p1 = buf;
    p2 = p1 + fread(buf, 1, 1 << 21, stdin);
    return p1 == p2 ? EOF : *p1++;
}
#define G gc

#ifndef ONLINE_JUDGE
#undef G
#define G getchar
#endif

template<class I>
inline void read(I &x) {
    x = 0; I f = 1; char c = G();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = G(); }
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = G(); }
    x *= f;
}

template<class I>
inline void write(I x) {
    if(x == 0) {putchar('0'); return;}
    I tmp = x > 0 ? x : -x;
    if(x < 0) putchar('-');
    int cnt = 0;
    while(tmp > 0) {
        buf1[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while(cnt > 0) putchar(buf1[--cnt]);
}

#define in(x) read(x)
#define outn(x) write(x), putchar('\n')
#define out(x) write(x), putchar(' ')

} using namespace io;

#define ll long long
const int N = 1000010;

int T, n, m;
int p[N], cnt, vis[N];
ll mu[N], S1[N], S2[N], S3[N];

void init() {
	mu[1] = 1;
	for(int i = 2; i < N; ++i) {
		if(!vis[i]) p[++cnt] = i, mu[i] = -1;
		for(int j = 1; j <= cnt && i * p[j] < N; ++j) {
			vis[i * p[j]] = 1;
			if(i % p[j] == 0) {
				mu[i * p[j]] = 0;
				break;
			}
			mu[i * p[j]] = -mu[i];
		}
	}
	for(int i = 1; i < N; ++i) {
		for(int j = i; j < N; j += i) {
			S3[j] += mu[i] * mu[j / i];
		}
	}
}

int main() {
	init(); read(T);
	while(T--) {
		read(n); read(m);
		for(int i = 1; i <= max(n, m); ++i) S1[i] = S2[i] = 0;
		if(n > m) swap(n, m);
		for(int i = 1; i <= n; ++i) 
			for(int j = i; j <= n; j += i) 
				S1[i] += mu[j];
		for(int i = 1; i <= m; ++i) 
			for(int j = i; j <= m; j += i) 
				S2[i] += mu[j];
		ll ans = 0;
		for(int i = 1; i <= n; ++i) {
			ans += S1[i] * S2[i] * S3[i];
		}
		outn(ans);
	}
}