【51nod】1742 开心的小Q

题解

我们由于莫比乌斯函数如果有平方数因子就是0，那么我们可以列出这样的式子

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

然后枚举倍数

\(\sum_{t = 1}^{n} \sum_{d = 1}^{\lfloor \frac{n}{t} \rfloor} (1 - |\mu(d)|)\)

\(\sum_{t = 1}^{n} F(\lfloor \frac{n}{t} \rfloor)\)

\(F(x)\)就表示1 - x有多少数有平方因子

可以用容斥得到

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

这个复杂度是\(n^{\frac{1}{3}}\)的，因为对于大于\(n^{\frac{1}{3}}\)的i，除数肯定小于\(n^{\frac{1}{3}}\)

然后我们的复杂度就是

\(O(\sqrt{n} + \sum_{i = 1}^{\sqrt{n}} (\frac{n}{i})^{\frac{1}{3}})\)可以解决问题

代码

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <vector>

#include <set>

#include <cmath>

#include <bitset>

#include <queue>

#define enter putchar('\n')

#define space putchar(' ')

//#define ivorysi

#define pb push_back

#define mo 974711

#define pii pair<int,int>

#define mp make_pair

#define fi first

#define se second

#define MAXN 200005

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 - '0' + c;

	c = getchar();

    }

    res = res * f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) out(x / 10);

    putchar('0' + x % 10);

}

int prime[100005],tot,mu[100005],M[100005];

bool nonprime[100005];

int F(int x) {

    int res = 0;

    for(int i = 1 ; i <= x / i ; ++i) {

	int r = sqrt(x / (x / (i * i)));

	if(x / ((r + 1) * (r + 1)) == x / (i * i)) ++r;

	if(x / (r * r) > x / (i * i)) --r;

	res += (x / (i * i)) * (M[r] - M[i - 1]);

	i = r;

    }

    return x - res;

}

int64 Solve(int x) {

    int64 res = 0;

    for(int i = 1 ; i <= x ; ++i) {

	int r = x / (x / i);

	res += 1LL * (r - i + 1) * F(x / i);

	i = r;

    }

    return res;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    mu[1] = 1;M[1] = 1;

    for(int i = 2 ; i <= 100000 ; ++i) {

	if(!nonprime[i]) {

	    prime[++tot] = i;

	    mu[i] = -1;

	}

	for(int j = 1 ; j <= tot ; ++j) {

	    if(prime[j] > 100000 / i) break;

	    nonprime[i * prime[j]] = 1;

	    if(i % prime[j] == 0) break;

	    else mu[i * prime[j]] = -mu[i];

	}

	M[i] = M[i - 1] + mu[i];

    }

    int a,b;

    read(a);read(b);

    out(Solve(b) - Solve(a - 1));

    enter;

}

巴特西

【51nod】1742 开心的小Q

题解

代码

最新文章

热门文章