【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。

[egin{aligned} &sum_{i=1}^nsum_{j=1}^nleft[frac{ij}{(i,j)}leq n ight]\ =&sum_{d=1}^nsum_{i=1}^{leftlfloorfrac nd ight floor}sum_{j=1}^{leftlfloorfrac nd ight floor}[(i,j)=1][ijdleq n]\ =&sum_{d=1}^nsum_{i=1}^{leftlfloorfrac nd ight floor}sum_{j=1}^{leftlfloorfrac nd ight floor}sum_{d'|(i,j)}mu(d')[ijdleq n]\ =&sum_{d=1}^nsum_{d'=1}^{leftlfloorfrac nd ight floor}mu(d')sum_{i=1}^{leftlfloorfrac n{dd'} ight floor}sum_{j=1}^{leftlfloorfrac n{dd'} ight floor}left[ijdd'^2leq n ight]\ =&sum_{d'=1}^{leftlfloorsqrt n ight floor}mu(d')sum_{d=1}^{leftlfloorfrac n{d'} ight floor}sum_{i=1}^{leftlfloorfrac n{dd'} ight floor}sum_{j=1}^{leftlfloorfrac n{dd'} ight floor}left[ijdleqleftlfloorfrac n{d'^2} ight floor ight]\ end{aligned} ]

枚举(leftlfloorsqrt n ight floor)个(d')，对于每个(d')，先假定(dleq ileq j)，枚举(leftlfloorleft(frac n{d'} ight)^{frac 13} ight floor)个(d)，(i)再从(d)枚举到(leftlfloorleft(frac n{dd'} ight)^{frac 12} ight floor)，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(Oleft(sumlimits_{d'=1}^{leftlfloorsqrt n ight floor}f(d') ight))。

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

[egin{aligned} &Oleft(sum_{d'=1}^{leftlfloorsqrt n ight floor}f(d') ight)\ =&Oleft(int_0^{sqrt n}left(frac n{x^2} ight)^{frac 23}dx ight)\ =&Oleft(n^{frac 23}int_0^{sqrt n}x^{-frac 43}dx ight)\ =&Oleft(n^{frac 23}left(left(-3sqrt{n}^{-frac 13} ight)-left(-3 imes0^{-frac 13} ight) ight) ight)\ =&Oleft(3n^{frac 23}-3n^{frac 12} ight)\ =&Oleft(n^{frac 23} ight) end{aligned} ]

所以总时间复杂度为(Oleft(n^{frac 23} ight))，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
", cal(b) - cal(a - 1));
	return 0;
}