题解

求积性函数的前缀和？杜教筛！

好了，然后我试着在这里推导一下式子

我们利用一个卷积就是
\(\mu * I = e\)
写成熟悉的形式就是
\([n = 1] = \sum_{d | n} \mu(d)\)
哎？和杜教筛有什么关系啊

$1 = \sum_{i = 1}^{n}[i = 1] = \sum_{i = 1}^{n} \sum_{d | i}\mu(d) = \sum_{i = 1}^{n} \sum_{d = 1}^{\lfloor \frac{n}{i} \rfloor} \mu(d) = \sum_{i = 1}^{n} M(\lfloor \frac{n}{i} \rfloor) \( \)M(n) = 1 - \sum_{i = 2}^{n} M(\lfloor \frac{n}{i} \rfloor)\( 这样的话，我们就可以只用\)\sqrt{n}\(个值算出\)M(n)$了
用一个hash记忆化搜索一下就好了= v =

代码

#include <bits/stdc++.h>
#define MAXN 10000005
//#define ivorysi
#define enter putchar('\n')
#define space putchar(' ')
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eps 1e-8
#define mo 974711
#define pii pair<int,int>
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);
}
struct node {
    int64 x,v;
    int next;
}E[1000005];
int head[mo + 5],sumE;
int64 a,b;
int prime[5000005],tot,M[MAXN],mu[MAXN];
bool nonprime[MAXN];
void add(int u,int64 x,int64 v) {
    E[++sumE].x = x;E[sumE].v = v;E[sumE].next = head[u];
    head[u] = sumE;
}
void Insert(int64 x,int64 v) {
    add(x % mo,x,v);
}
int64 Query(int64 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 1e18;
}
int64 f(int64 x) {
    if(x <= 10000000) return M[x];
    int64 c = Query(x); 
    if(c != 1e18) return c;
    int64 res = 0;
    for(int64 i = 2 ; i <= x ; ++i) {
	int64 r = x / (x / i);
	res += (r - i + 1) * f(x / i);
	i = r;
    }
    res = 1 - res;
    Insert(x,res);
    return res;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    mu[1] = 1;M[1] = 1;
    for(int i = 2 ; i <= 10000000 ; ++i) {
	if(!nonprime[i]) {
	    mu[i] = -1;
	    prime[++tot] = i;
	}
	for(int j = 1 ; j <= tot ; ++j) {
	    if(prime[j] > 10000000 / i) break;
	    nonprime[prime[j] * i] = 1;
	    if(i % prime[j] == 0) mu[i * prime[j]] = 0;
	    else mu[i * prime[j]] = -mu[i];
	}
	M[i] = M[i - 1] + mu[i];
    }
    read(a);read(b);
    out(f(b) - f(a - 1));enter;
    return 0;
}