【51Nod 1222】最小公倍数计数

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1222

求\([a,b]\)中的个数转化为求\([1,b]\)中的个数减去\([1,a)\)中的个数。

\[\begin{aligned}
&\sum_{i=1}^n\sum_{j=1}^n\left[\frac{ij}{(i,j)}\leq n\right]\\
=&\sum_{d=1}^n\sum_{i=1}^{\left\lfloor\frac nd\right\rfloor}\sum_{j=1}^{\left\lfloor\frac nd\right\rfloor}[(i,j)=1][ijd\leq n]\\
=&\sum_{d=1}^n\sum_{i=1}^{\left\lfloor\frac nd\right\rfloor}\sum_{j=1}^{\left\lfloor\frac nd\right\rfloor}\sum_{d'|(i,j)}\mu(d')[ijd\leq n]\\
=&\sum_{d=1}^n\sum_{d'=1}^{\left\lfloor\frac nd\right\rfloor}\mu(d')\sum_{i=1}^{\left\lfloor\frac n{dd'}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac n{dd'}\right\rfloor}\left[ijdd'^2\leq n\right]\\
=&\sum_{d'=1}^{\left\lfloor\sqrt n\right\rfloor}\mu(d')\sum_{d=1}^{\left\lfloor\frac n{d'}\right\rfloor}\sum_{i=1}^{\left\lfloor\frac n{dd'}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac n{dd'}\right\rfloor}\left[ijd\leq\left\lfloor\frac n{d'^2}\right\rfloor\right]\\
\end{aligned}
\]

枚举\(\left\lfloor\sqrt n\right\rfloor\)个\(d'\)，对于每个\(d'\)，先假定\(d\leq i\leq j\)，枚举\(\left\lfloor\left(\frac n{d'}\right)^{\frac 13}\right\rfloor\)个\(d\)，\(i\)再从\(d\)枚举到\(\left\lfloor\left(\frac n{dd'}\right)^{\frac 12}\right\rfloor\)，最后计算\(j\)的个数，乘上一个排列数即可。

\(f(x)\)表示对每个\(d'\)计算的复杂度。

时间复杂度为\(O\left(\sum\limits_{d'=1}^{\left\lfloor\sqrt n\right\rfloor}f(d')\right)\)。

\[\begin{aligned}
f(x)=&\int_{0}^{\left(\frac n{x^2}\right)^{\frac 13}}\left(\left(\frac n{x^2i}\right)^{\frac 12}-i\right)di\\
=&\left(\frac n{x^2}\right)^{\frac 12}\int_0^{\left(\frac n{x^2}\right)^{\frac 13}}i^{-\frac 12}di-\int_0^{\left(\frac n{x^2}\right)^{\frac 13}}idi\\
=&\left(\frac n{x^2}\right)^{\frac 12}\left(2\left({\left(\frac n{x^2}\right)^{\frac 13}}\right)^{\frac 12}-2\right)-\left(\frac 12\left({\left(\frac n{x^2}\right)^{\frac 13}}\right)^2\right)\\
=&\frac 34\left(\frac n{x^2}\right)^{\frac 23}-2\left(\frac n{x^2}\right)^{\frac 12}
\end{aligned}
\]

\[\begin{aligned}
&O\left(\sum_{d'=1}^{\left\lfloor\sqrt n\right\rfloor}f(d')\right)\\
=&O\left(\int_0^{\sqrt n}\left(\frac n{x^2}\right)^{\frac 23}dx\right)\\
=&O\left(n^{\frac 23}\int_0^{\sqrt n}x^{-\frac 43}dx\right)\\
=&O\left(n^{\frac 23}\left(\left(-3\sqrt{n}^{-\frac 13}\right)-\left(-3\times0^{-\frac 13}\right)\right)\right)\\
=&O\left(3n^{\frac 23}-3n^{\frac 12}\right)\\
=&O\left(n^{\frac 23}\right)
\end{aligned}
\]

所以总时间复杂度为\(O\left(n^{\frac 23}\right)\)，好神奇啊qwq

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

const int sq = 316228;

bool notp[sq + 1];
int mu[sq + 1], prime[sq], num = 0;

void Euler_shai() {
	mu[1] = 1;
	for (int i = 2; i <= sq; ++i) {
		if (!notp[i]) prime[++num] = i, mu[i] = -1;
		for (int j = 1; j <= num && prime[j] * i <= sq; ++j) {
			notp[prime[j] * i] = true;
			if (i % prime[j] == 0)
				break;
			mu[prime[j] * i] = -mu[i];
		}
	}
}

ll a, b;

ll cal(ll n) {
	ll ret = 0;
	for (int d = 1; 1ll * d * d <= n; ++d)
		if (mu[d]) {
			ll up = n / d / d, r = 0;
			for (int i = 1; 1ll * i * i * i <= up; ++i) {
				ll up2 = up / i;
				r += (up2 / i - i) * 3 + 1;
				for (int j = i + 1; 1ll * j * j <= up2; ++j)
					r += (up2 / j - j) * 6 + 3;
			}
			mu[d] > 0 ? ret += r : ret -= r;
		}
	return (ret + n) >> 1;
}

int main() {
	scanf("%lld%lld", &a, &b);
	Euler_shai();
	printf("%lld\n", cal(b) - cal(a - 1));
	return 0;
}