P3327 [SDOI2015]约数个数和

推导过程

求${\sum }_{i=1}^n{\sum }_{j=1}^{m}d(ij)$

$$=\sum _{i=1}^n\sum _{j=1}^m\sum _{x\mid i}\sum _{y\mid j}gcd(x,y)==1$$

改成枚举$x,y$，

$$=\sum _{i=1}^n\sum _{j=1}^m\sum _{i\mid x}\sum _{j\mid y}gcd(i,j)==1$$

$$=\sum _{i=1}^n\sum _{j=1}^m\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{j}\rfloor (gcd(i,j)==1)$$

$$=\sum _{i=1}^n\sum _{j=1}^m\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{j}\rfloor \sum _{d\mid (gcd(i,j))}\mu (d)$$

$$\sum _{d=1}^n\mu (d)\sum _{i=1}^{\frac{n}{d}}\lfloor \frac{n}{di}\rfloor \sum _{j=1}^{\frac{m}{d}}\lfloor \frac{m}{dj}\rfloor$$

最后我们只要预处理一下所有$\frac{n}{d}\frac{m}{d}$的整除分块即可，整体复杂度$n\sqrt{n}$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 5e4 + 10;bool st[N];vector<int> prime;int n, m, mu[N];ll sum[N];void mobius() {st[0] = st[1] = mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime.pb(i);mu[i] = -1;}for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] += mu[i - 1];for(ll l = 1, r; l <= i; l = r + 1) {r = i / (i / l);sum[i] += (i / l) * (r - l + 1);}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);mobius();int T = read();while(T--) {ll n = read(), m = read(), ans = 0;if(n > m) swap(n, m);for(ll l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans += 1ll * (mu[r] - mu[l - 1]) * sum[n / l] * sum[m / l];}printf("%lld\n", ans);}return 0;
}