# 【51Nod 1222】最小公倍数计数

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1222
$$[a,b]$$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}

$$f(x)$$表示对每个$$d'$$计算的复杂度。

\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}

#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;
}

posted @ 2017-04-27 17:19  abclzr  阅读(289)  评论(0编辑  收藏  举报