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(nlog n)暴力算...

[egin{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) ight) left( sum_{j} ^ {frac{m}{T}} mu(jT) ight)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('
')
#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);
	}
}