【51Nod 1190】最小公倍数之和 V2

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1190

$$\begin{aligned} &\sum_{i=a}^b\frac{ib}{(i,b)}\\ =&b\sum_{i=a}^b\frac i{(i,b)}\\ =&b\sum_{d|b}\sum_{i=a}^b[d|i]\left[\left(\frac id,\frac bd\right)=1\right]\frac id\\ =&b\sum_{d|b}\sum_{i=\left\lceil\frac ad\right\rceil}^{\frac bd}\left[\left(i,\frac bd\right)=1\right]i\\ =&b\sum_{d|b}\sum_{i=\left\lceil\frac ad\right\rceil}^{\frac bd}i\sum_{d'|i,d'|\frac bd}\mu(d')\\ =&b\sum_{d|b}\sum_{d'|\frac bd}\mu(d')\sum_{i=\left\lceil\frac {a}{dd'}\right\rceil}^{\frac{b}{dd'}}id'\\ =&b\sum_{T|b}\sum_{d|T}\mu(d)\sum_{i=\left\lceil\frac aT\right\rceil}^{\frac bT}id\\ =&b\sum_{T|b}\frac{\left(\left\lceil\frac aT\right\rceil+\frac bT\right)\left(\frac bT-\left\lceil\frac aT\right\rceil+1\right)}{2}\sum_{d|T}\mu(d)d \end{aligned}$$

$\sum\limits_{d|T}\mu(d)d=\prod\left(1-p_i\right)$，只要确定T的质因子就可以确定$\sum\limits_{d|T}\mu(d)d$的值。
如果循环枚举T找b的约数，无法快速计算T的质因子。
可以dfs枚举b的约数T，动态计算$\sum\limits_{d|T}\mu(d)d$的值。
时间复杂度$O\left(T\sqrt n\right)$。

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

const int N = 100000;
const int p = 1000000007;
const int ni2 = 500000004;

bool notp[N];
int a, b, tot, P[N], c[N], num = 0, prime[N];

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

void pre(int x) {
	tot = 0;
	for (int i = 1, pi = 2; i <= num && pi * pi <= x; pi = prime[++i])
		if (x % pi == 0) {
			P[++tot] = pi; c[tot] = 0;
			while (x % pi == 0) x /= pi, ++c[tot];
		}
	if (x > 1)
		P[++tot] = x, c[tot] = 1;
}

int ans;

void dfs(int tmp, int T, int f) {
	if (tmp > tot) {
		int l = a / T, r = b / T;
		if (a % T) ++l;
		(ans += 1ll * (l + r) * (r - l + 1) % p * ni2 % p * f % p) %= p;
		return;
	}
	dfs(tmp + 1, T, f);
	int tt = T, ff = 1ll * f * (1 - P[tmp] + p) % p;
	for (int i = 1; i <= c[tmp]; ++i) {
		tt *= P[tmp];
		dfs(tmp + 1, tt, ff);
	}
}

int main() {
	Euler_shai();
	
	int T; scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &a, &b);
		pre(b);
		ans = 0;
		dfs(1, 1, 1);
		printf("%lld\n", 1ll * b * ans % p);
	}
	
	return 0;
}