mobius初步

求 $\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (gcd(i, j) == 1)$

我们引入一个知识 $\sum_{d \mid n} \mu(d) = (n == 1)$

所以 $gcd(i, j) = \sum_{d \mid gcd(i, j)} \mu(d)$

对上式进行化简：

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{m} \sum_{d\mid gcd(i, j)} \mu(d)$

$= \sum_{d \mid n} \mu(d) \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1}^{\frac{m}{d}}1$

$\sum_{d = 1} ^{n} \mu(d) \frac{n}{d} \frac{m}{d}$

最后利用整除分块求得最后的答案
求 $\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} (gcd(i, j) == k)$

$= \sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{m}{d}} (gcd(i, j) == 1)$

另 $n' = \frac{n}{k}, m' = \frac{m}{k}$ ，所以如上式：

P3455 [POI2007]ZAP-Queries

上面式子的模板题，就不过多讲解了，直接上代码

/*
  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 x = 0, f = 1; 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 x * f;
}

const int N = 5e4 + 10;

int mu[N], a, b, d;

bool st[N];

vector<int> prime;

void mobius() {
    st[1] = st[0] = 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) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) mu[i] += mu[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);
    mobius();
    int T = read();
    while(T--) {
        a = read(), b = read(), d = read();
        int n = a /= d, m = b /= d;
        if(n > m) swap(n, m);
        ll ans = 0;
        for(int l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += 1ll * (mu[r] - mu[l - 1]) * (n / l) * (m / l);
        }
        printf("%lld\n", ans);
    }
	return 0;
}

P2522 [HAOI2011]Problem b

也是是模板题，只是稍加改动，类似于前缀和的取法，最后得到我们的答案。

/*
  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 x = 0, f = 1; 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 x * f;
}

const int N = 5e4 + 10;

int mu[N], a, b, c, d, k;

bool st[N];

vector<int> prime;

void mobius() {
    st[1] = st[0] = 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) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) mu[i] += mu[i - 1];
}

ll solve(int a, int b, int d) {
    int n = a /= d, m = b /= d;
    if(n > m) swap(n, m);
    ll ans = 0;
    for(int l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans += 1ll * (mu[r] - mu[l - 1]) * (n / l) * (m / l);
    }
    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();
    int T = read();
    while(T--) {
        a = read(), b = read(), c = read(), d = read(), k = read();
        ll ans = solve(b, d, k) + solve(a - 1, c - 1, k) - solve(a - 1, d, k) - solve(b, c - 1, k);
        printf("%lld\n", ans);
    }
	return 0;
}

$\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} ij(gcd(i, j) == k)$

$= k ^ 2\sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{m}{k}}ij(gcd(i, j) == 1)$

$= k ^ 2\sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{m}{k}}ij\sum_{d\mid gcd(i, j)} \mu(d)$

$= k ^ 2\sum_{d = 1} ^{\frac{n}{k}} \mu(d)\sum_{i = 1} ^{\frac{n}{k}} \sum_{j = 1} ^{\frac{m}{k}}ij$

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

$\sum_{i = 1} ^{\frac{n}{kd}} \sum_{j = 1} ^{\frac{m}{kd}}ij$ 这一项可以通过整除分块得到，最后再统计一遍答案就行。

P3455 [POI2007]ZAP-Queries

P2522 [HAOI2011]Problem b

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