YY的GCD_牛客博客

YY的GCD

推式子

$\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} [gcd(i, j) \in primes]\\ \sum_{d \in primes}\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} [gcd(i, j) = 1]\\ \sum_{d \in primes} \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}\\$

到这一步可以用min_25筛出区间质数个数，然后再整除分块来写，

但是显然复杂度有点高，继续化简。 $t = kd\\ \sum_{t = 1} ^{n} \frac{n}{t} \frac{m}{t} \sum_{d \mid t, d \in primes} \mu(\frac{t}{d})\\$

整体复杂度 $O(NlogN + T sqrt(N))$

代码

/*
  Author : lifehappy
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 1e7 + 10;

int mu[N], prime[N], cnt, n, m;

ll sum[N];

bool st[N];

void init() {
    st[0] = st[1] = mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[++cnt] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i <= cnt; i++) {
        for(int j = 1; prime[i] * j < N; j++) {
            sum[j * prime[i]] += mu[j];
        }
    }
    for(int i = 1; i < N; i++) sum[i] += sum[i - 1];
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d", &n, &m);
        if(n > m) swap(n, m);
        ll ans = 0;
        for(ll l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += 1ll * (sum[r] - sum[l - 1]) * (n / l) * (m / l);
        }
        printf("%lld\n", ans);
    }
    return 0;
}