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$n^{\prime }=\frac{n}{k},m^{\prime }=\frac{m}{k}$，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 | 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 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 | 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 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。

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