P1829 [国家集训队]Crash的数字表格（推了好久的mobius反演）

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 $sum_{k = 1} ^{\frac{n}{d}}k ^ 2 \mu(k)$，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 | 1

using 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;
}