hdu 5382 GCD?LCM! - 莫比乌斯反演

题目传送门

　　传送门I

　　传送门II

题目大意

　　设$F(n) = \sum_{i = 1}^{n}\sum_{j = 1}^{n}\left [ [i, j] + (i, j) \geqslant n \right ]$，求$\sum_{i = 1}^{n} F(i)$。

　　考虑设$f(n) = \sum_{i = 1}^{n}\sum_{i = 1}^{n}\left[ \left[i,j \right ] + (i, j) = n \right ]$，那么有$F(n) = F(n - 1) - f(n - 1) + 2n - 1$。（因为$[i, n]\geqslant n$）

　　显然$[i, j] = k(i, j)$，那么我们可以枚举最大公约数。

　　$f(n)=\sum_{d \mid n}\sum_{i = 1}^{n}\sum_{j = 1}^{n}[(i, j) = 1][[id, jd] +d = n]\\=\sum_{d \mid n}\sum_{i = 1}^{n}\sum_{j = 1}^{n}[(i, j) = 1][ijd +d = n]\\=\sum_{d\mid n}\sum_{id | (n - d)}[(i, \frac{n - d}{di}) = 1]$

　　然后就是套路了。

　　$f(n)=\sum_{d\mid n}\sum_{id | (n - d)}[(i, \frac{n - d}{di}) = 1]\\=\sum_{d|n}\sum_{id|(n-d)}\sum_{e|i \wedge edi\mid (n - d)}\mu(e)\\=\sum_{d|n}\sum_{de|(n - d)}\mu(e)\left( \sum_{e|i\wedge ide\mid(n - d)}1\right )\\=\sum_{d|n}\sum_{e^2d|(n - d)}\mu(e)\sigma_0\left(\frac{n-d}{e^2d}\right)$

　　设$g(n) = \sum_{e^2|n}\mu(e)\sigma_0(\frac{n}{e^2})$，显然$g(n)$是可以$O(n\log n)$预处理的。考虑$i$在$1, 2,\cdots, n$中作为$\left \lfloor \frac{n}{i} \right \rfloor$个数的因子，然后枚举的总的因子的个数就是一个调和级数的和。对于$f(n)$也可以这样处理。然后就过了。

Code

 /**
  * hdu
  * Problem#5382
  * Accepted
  * Time: 655ms
  * Memory: 33992k
  */
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <ctime>
 using namespace std;
 typedef bool boolean;

 const int N = 1e6 + , M = ;

 int add(int a, int b) {
     a += b;
     if (a < )
         a += M;
     if (a >= M)
         a -= M;
     return a;
 }

 int T, n;
 int f[N], F[N], S[N];
 int sl[N], mp[N], mu[N];
 int rou[N], g[N];
 boolean vis[N];
 int num, pri[];

 int tp;
 int fac[];

 void dfs(int x, int d) {
     if (x == ) {
         fac[tp++] = d;
         return;
     }
     int p = mp[x], link = sl[x];
     dfs(sl[x], d);
     while (!(x % p))
         x /= p, d *= p, dfs(link, d);
 }

 void dfs1(int x, int d) {
     if (x == ) {
         fac[tp++] = d;
         return;
     }
     int p = mp[x], a = ;
     while (!(x % p))
         x /= p, a++;
     a = a >> ;
     for (int i = , pro = ; i <= a; i++, pro = pro * p)
         dfs1(x, pro * d);
 }

 inline void prepare() {
     mu[] = , rou[] = ;
     for (int i = ; i < N; i++) {
         if (!vis[i])
             pri[num++] = mp[i] = i, mu[i] = -, rou[i] = , sl[i] = ;
         for (int j = , x; j < num && pri[j] * i < N; j++) {
             x = pri[j] * i, vis[x] = true;
             if (i % pri[j])
                 mu[x] = -mu[i], mp[x] = pri[j], sl[x] = i, rou[x] = rou[i] << ;
             else {
                 mu[x] = , mp[x] = pri[j], sl[x] = sl[i], rou[x] = rou[x / pri[j]] + rou[sl[x]];
                 break;
             }
         }
     }

     g[] = ;
     for (int i = ; i < N; i++) {
         tp = , dfs1(i, );
         for (int j = , x, d; j < tp; j++) {
             x = fac[j];
             d = i / x / x;
             if (d * x * x == i)
                 g[i] = add(g[i], mu[x] * rou[i / x / x]);
         }
     }

     f[] = ;
     int cnt = ;
     for (int i = ; i < N; i++) {
         tp = , dfs(i, );
         cnt += tp;
         for (int j = , x; j < tp; j++) {
             x = fac[j];
             f[i] = add(f[i], g[(i - x) / x]);
         }
     }

     F[] = ;
     for (int i = ; i < N; i++)
         F[i] = add(add(F[i - ], -f[i - ]),  * i - );
     S[] = ;
     for (int i = ; i < N; i++)
         S[i] = add(S[i - ], F[i]);
 }

 inline void solve() {
     scanf("%d", &n);
     printf("%d\n", S[n]);
 }

 int main() {
     prepare();
     scanf("%d", &T);
     while (T--)
         solve();
     return ;
 }