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题目链接：ybtoj高效进阶 21177

题目大意

给你求 ∑i=1~n∑j=1~mφ(gcd(i,j))。

思路

重新写好看点：
\(\sum\limits_{i=1}^n\sum\limits_{j=1}^m\varphi(\gcd(i,j))\)
\(\sum\limits_{d}\varphi(d)\sum\limits_{i=1}^n\sum\limits_{j=1}^m[\gcd(i,j)=d]\)
\(\sum\limits_{d}\varphi(d)\sum\limits_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum\limits_{j=1}^{\left\lfloor\frac{m}{d}\right\rfloor}[\gcd(i,j)=1]\)
\(\sum\limits_{d}\varphi(d)\sum\limits_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum\limits_{j=1}^{\left\lfloor\frac{m}{d}\right\rfloor}\sum\limits_{p|gcd(i,j)}\mu(p)\)
\(\sum\limits_{d}\varphi(d)\sum\limits_{p}\sum\limits_{i=1,i|p}^{\left\lfloor\frac{n}{d}\right\rfloor}\sum\limits_{j=1,j|p}^{\left\lfloor\frac{m}{d}\right\rfloor}\mu(p)\)
\(\sum\limits_{d}\varphi(d)\sum\limits_{p}\mu(p)\cdot\left\lfloor\dfrac{n}{dp}\right\rfloor\cdot\left\lfloor\dfrac{m}{dp}\right\rfloor\)

然后设 \(T=pd\)，然后有：
\(\sum\limits_{d}\varphi(d)\sum\limits_{p}\mu(p)\cdot\left\lfloor\dfrac{n}{T}\right\rfloor\cdot\left\lfloor\dfrac{m}{T}\right\rfloor\)
\(\sum\limits_{T=1}^{\min(n,m)}\left\lfloor\dfrac{n}{T}\right\rfloor\cdot\left\lfloor\dfrac{m}{T}\right\rfloor\cdot(\sum\limits_{d|T}\varphi(d)\mu(\dfrac{T}{d}))\)

然后你会发现右边就是一个狄利克雷卷积。
那我们设 \(g=\varphi*\mu\)（积性函数，因为是两个积性函数相乘），然后式子就是：
\(\sum\limits_{T=1}^{\min(n,m)}\left\lfloor\dfrac{n}{T}\right\rfloor\cdot\left\lfloor\dfrac{m}{T}\right\rfloor\cdot g(d)\)

然后你发现左边两个可以数论分块，然后你考虑右边的能不能求前缀和。
看到是积性函数，考虑用杜教筛，设 \(S(n)=\sum\limits_{i=1}^ng(i)\)。

然后你就要找一个合适的 \(f\)，不难看出可以用 \(I*I=d\)。
然后 \(g*d=\varphi*\mu*I*I=(\varphi*I)*(\mu*I)=id*\epsilon=id\)。

然后就有 \(d(1)S(n)=\sum\limits_{i=1}^nid(i)-\sum\limits_{i=2}^nd(i)S(\left\lfloor\dfrac{n}{i}\right\rfloor)=\dfrac{n(n+1)}{2}-\sum\limits_{i=2}^nd(i)S(\left\lfloor\dfrac{n}{i}\right\rfloor)\)

然后 \(d(i)\) 的前缀和可以数论分块，小于 \(n^{\frac{2}{3}}\) 的预处理前缀和，大于的直接根号求。

然后小小卡个常即可。

代码

#include<map>
#include<cstdio>
#define ll long long
#define mo 1000000007

using namespace std;

const int Maxn = 2000001;
int n, m;
ll d[Maxn + 1], g[Maxn + 1], ans;
int low[Maxn + 1], phi[Maxn + 1], prime[Maxn + 1];
map <int, ll> ans_g;

void init() {//杜教筛的预处理
	g[1] = 1; phi[1] = 1; low[1] = 1;
	for (int i = 2; i <= Maxn; i++) {
		if (!low[i]) phi[i] = i - 1, low[i] = i, prime[++prime[0]] = i, g[i] = i - 2;
		for (int j = 1; j <= prime[0] && i * prime[j] <= Maxn; j++) {
			if (i % prime[j]) low[i * prime[j]] = prime[j], phi[i * prime[j]] = phi[i] * (prime[j] - 1), g[i * prime[j]] = g[i] * g[prime[j]];
				else {
					low[i * prime[j]] = low[i] * prime[j], phi[i * prime[j]] = phi[i] * prime[j];
					if (low[i] != i) g[i * prime[j]] = g[i / low[i]] * g[low[i] * prime[j]];
						else g[i * prime[j]] = phi[i * prime[j]] + phi[i] * (-1);
					break;
				}
		}
	}
	for (int i = 1; i <= Maxn; i++)
		for (int j = 1; i * j <= Maxn; j++)
			d[i * j]++;
	for (int i = 1; i <= Maxn; i++)
		d[i] += d[i - 1], g[i] += g[i - 1];
}

ll sum_d(ll n) {//数论分块
	if (n <= Maxn) return d[n];
	ll re = 0;
	for (ll l = 1, r; l <= n; l = r + 1) {
		r = n / (n / l);
		re += (r - l + 1) * (n / l);
	}
	return re;
}

ll sum_g(ll n) {//杜教筛
	if (n <= Maxn) return g[n];
	if (ans_g[n]) return ans_g[n];
	ll re = n * (n + 1) / 2;
	for (ll l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		re -= (sum_d(r) - sum_d(l - 1)) * sum_g(n / l);
	}
	return ans_g[n] = re;
}

int main() {
//	freopen("mesh.in", "r", stdin);
//	freopen("mesh.out", "w", stdout);
	
	scanf("%d %d", &n, &m);
	
	init();
	
	for (int l = 1, r; l <= n && l <= m; l = r + 1) {//数论分块
		r = min(n / (n / l), m / (m / l));
		ans = (ans + 1ll * (n / l) * (m / l) * (sum_g(r) - sum_g(l - 1))) % mo;
	}
	
	printf("%lld", ans);
	
	return 0;
}