P1829 [国家集训队]Crash的数字表格 / JZPTAB

推导过程

$$\sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)$$

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

$$=\sum _{d=1}^n\frac{1}{d}\sum _{i=1}^n\sum _j^{m}ij(gcd(i,j)==d)$$

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

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

$$=\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}{k}^2\mu (k)\sum _{i=1}^{\frac{n}{kd}}i\sum _{j=1}^{\frac{m}{kd}}j$$

$$=\sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}{k}^2\mu (k)\frac{\lfloor \frac{n}{kd}\rfloor (\lfloor \frac{n}{kd}\rfloor +1)}{2}\frac{\lfloor \frac{m}{kd}\rfloor (\lfloor \frac{m}{kd}\rfloor +1)}{2}$$

所以我们只要预处理出$su{m}_{k=1}^{\frac{n}{d}}{k}^2\mu (k)$，然后再进行两次分块即可。

整体复杂度$\sqrt{n}\sqrt{n}=n$还是线性$O(n)$的。

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 = 1e7 + 10, mod = 20101009;bool st[N];vector<int> prime;int n, m;
ll mu[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] + mu[i] *  i % mod * i % mod + mod) % mod;}
}ll calc1(ll l, ll r) {return (l + r) * (r - l + 1) / 2 % mod;
}ll calc2(ll n, ll m) {ll 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 = (ans + (mu[r] - mu[l - 1] + mod) % mod * calc1(1, n / l) % mod * calc1(1, m / l) % mod + mod) % mod;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);mobius();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 = (ans + calc1(l, r) * calc2(n / l, m / l) % mod) % mod;}printf("%lld\n", ans);return 0;
}