【莫比乌斯反演】10.30破译密码

初涉的话先留坑吧

题目大意

$\sum_{i_1}^{a_1}\sum_{i_2}^{a_2}\cdots\sum_{i_m}^{a_m}(i_1,i_2,\cdots,i_m)$

$a_i<=1e6,2=<m<=10$

题目分析

首先寄存两篇比较好的博客：

1.铃悬的数学小讲堂——狄利克雷卷积与莫比乌斯反演

2.[学习笔记]莫比乌斯反演常见模型

这个问题可以推广至一类模型：$\sum_{i=1}^n\sum_{j=1}^mf[\gcd(i,j)]$.

该模型的推论是$原式=\sum_{u=1}^{\min(n,m)}\lfloor\frac{n}{u}\rfloor\lfloor\frac{m}{u}\rfloor\sum_{d|u}f[d]\mu(\frac{u}{d})$

注意到在本题中$f=id$，那么也就是说$原式=\sum_{u=1}^{\min(n,m)}\lfloor\frac{n}{u}\rfloor\lfloor\frac{m}{u}\rfloor \varphi(u)$.

因此先一遍线性筛求$\varphi$的前缀和，再数论分块做$\lfloor\frac{n}{u}\rfloor$这一部分。

本题对模型的转化还不算太深（算是比较裸的反演）

 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 const int maxn = 13;
 4 const int MO = 1e9+7;
 5 const int TOP = 1000000;
 6 const int maxPri = 80035;
 7 const int maxNum = 1000035;
 8 
 9 int T,n,mn,a[maxn],pr[maxPri];
10 ll phi[maxNum],ans,tmp;
11 bool vis[maxNum];
12 
13 int read()
14 {
15     char ch = getchar();
16     int num = 0;
17     bool fl = 0;
18     for (; !isdigit(ch); ch=getchar())
19         if (ch=='-') fl = 1;
20     for (; isdigit(ch); ch=getchar())
21         num = (num<<1)+(num<<3)+ch-48;
22     if (fl) num = -num;
23     return num;
24 }
25 void init()
26 {
27     phi[1] = 1;
28     for (int i=2; i<=TOP; i++)
29     {
30         if (!vis[i]) pr[++pr[0]] = i, phi[i] = i-1;
31         for (int j=1; (j<=pr[0])&&(pr[j]*i<=TOP); j++)
32         {
33             vis[pr[j]*i] = 1, phi[pr[j]*i] = phi[i]*pr[j];
34             if (i%pr[j]==0) break;
35             phi[pr[j]*i] = phi[i]*(pr[j]-1);
36         }
37     }
38     for (int i=2; i<=TOP; i++) phi[i] = (phi[i]+phi[i-1])%MO;
39 }
40 int main()
41 {
42     freopen("gcd.in","r",stdin);
43     freopen("gcd.out","w",stdout);
44     T = read(), init();
45     while (T--)
46     {
47         n = read(), ans = 0, mn = 0x3f3f3f3f;
48         for (int i=1; i<=n; i++)
49             a[i] = read(), mn = mn>a[i]?a[i]:mn;
50         for (int i=1, j=0; i<=mn; i=j+1)
51         {
52             j = mn, tmp = 1;
53             for (int k=1; k<=n; k++) j = std::min(j, a[k]/(a[k]/i));
54             for (int k=1; k<=n; k++) tmp = tmp*(a[k]/i)%MO;
55             ans = (ans+(phi[j]-phi[i-1]+MO)*tmp%MO)%MO;
56         }
57         printf("%lld\n",ans);
58     }
59     return 0;
60 }