【51nod】2026 Gcd and Lcm

题解

话说LOJ说我今天宜学数论= =看到小迪学了杜教筛去蹭了一波小迪做的题

标解的杜教筛的函数不懂啊，怎么推的毫无思路= =

所以写了个复杂度稍微高一点的？？

首先，我们发现f是个积性函数，那么我们就有……

\(\prod_{i = 1}^{k}f(p_{i}^{a_{i}})\)

我们发现，对于每个质因子，gcd是取较小值，lcm取较大值

\(f(lcm(x,y)) * f(gcd(x,y)) = \prod_{i = 1}^{k} f(p_{i}^{max(a_{i},b_{i}) + min(a_{i},b_{i})})\)

\(max(a,b) + min(a,b) = a + b\)

那么就有

\(f(lcm(x,y)) * f(gcd(x,y)) = \prod_{i = 1}^{k} f(p_{i}^{max(a_{i},b_{i}) + min(a_{i},b_{i})}) = f(x) * f(y)\)

所以我们只要求出\([\sum_{i = 1}^{n} f(i)]^2\)就是答案了！

怎么求呢

\(S(n) = \sum_{i = 1}^{n}\sum_{d | i} \mu(d)\cdot d\)

\(S(n) = \sum_{d = 1}^{n}\sum_{d | i} \mu(d)\cdot d\)

\(S(n) = \sum_{d = 1}^{n} \mu(d)\cdot d \cdot \lfloor \frac{n}{d} \rfloor\)

我们可以数论分块处理\(\lfloor \frac{n}{d} \rfloor\)

那么我们考虑计算\(\sum_{d = 1}^{n} \mu(d)\cdot d\)

我们发现这个函数卷上一个\(Id(x)\)等于\(e\)

\(\sum_{i = 1}^{n} [i = 1] = \sum_{i = 1}^{n} \sum_{d |i} \mu(d) \cdot d \cdot \frac{i}{d} = \sum_{k = 1}^{n} k \sum_{d}^{\frac{n}{k}} \mu(d) \cdot d\)

所以最后就是

\(S(n) = 1 - \sum_{i = 2}^{n} S(\lfloor \frac{n}{i}\rfloor)\)

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 1000000
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
	    while(c < '0' || c > '9') {
			if(c == '-') f = -1;
			c = getchar();
	    }
	    while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
		out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
const int MOD = 1000000007;
struct node {
	int x,v,next;
}E[2000006];
int head[mo + 5],sumE;
int prime[MAXN + 5],tot,S[MAXN + 5],mu[MAXN + 5];
bool nonprime[MAXN + 5];
int inc(int a,int b) {
	a = a + b;
	if(a >= MOD) a -= MOD;
	return a;
}
void add(int u,int x,int v) {
	E[++sumE].x = x;E[sumE].v = v;E[sumE].next = head[u];
	head[u] = sumE;
}
void Insert(int x,int v) {
	add(x % mo,x,v);
}
int Query(int x) {
	int u = x % mo;
	for(int i = head[u] ; i ; i = E[i].next) {
		if(E[i].x == x) return E[i].v;
	}
	return -1;
}
int f(int x) {
	if(x <= MAXN) return S[x];
	int c = Query(x);
	if(c != -1) return c;
	int res = 0;
	for(int i = 2 ; i <= x ; ++i) {
		int r = x / (x / i);
		res = inc(res,1LL * (r - i + 1) * (r + i) / 2 % MOD * f(x / i) % MOD);
		i = r;
	}
	res = inc(1,MOD - res);
	Insert(x,res);
	return res;
}
void Solve() {
	mu[1] = 1;S[1] = 1;
	for(int i = 2 ; i <= MAXN ; ++i) {
		if(!nonprime[i]) {
			mu[i] = -1;
			prime[++tot] = i;
		}
		for(int j = 1 ; j <= tot ; ++j) {
			if(prime[j] > MAXN / i) break;
			nonprime[i * prime[j]] = 1;
			if(i % prime[j] == 0) break;
			else mu[i * prime[j]] = -mu[i];
		}
		S[i] = (S[i - 1] + mu[i] * i + MOD) % MOD;
	}
	read(N);
	int res = 0;
	for(int i = 1 ; i <= N ; ++i) {
		int r = N / (N / i);
		res = inc(1LL * (f(r) + MOD - f(i - 1)) * (N / i) % MOD,res);
		i = r;
	}
	out(1LL * res * res % MOD);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
   	Solve();
    return 0;
}