Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the rest of the world is fighting for the Iron Throne, he is going to get ready for the attack of the White Walkers.

He has created a method to know how strong his army is. Let the i

Your task is to find the strength of his army. As the number may be very large, you have to print it modulo 1000000007 (109 + 7).

Greatest common divisor (gcd) of a sequence of integers is the maximum possible integer so that each element of the sequence is divisible by it.

The first line contains integer n (1 ≤ n ≤ 200000) — the size of the army.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000000) — denoting the strengths of his soldiers.

Print one integer — the strength of John Snow‘s army modulo 1000000007 (109 + 7).

3
3 3 1

12

4
2 3 4 6

39

In the first sample the clans are {1}, {2}, {1, 2} so the answer will be 1·3 + 1·3 + 2·3 = 12

　　題目大意 給定n，集合A，設表示把這個集合內的所有數求最大公約數的結果，求。

　　根據常用套路，套一個循環去枚舉gcd的結果，然後再求系數，於是有

　　現在設,於是有

　　現在考慮求f(i)。可以想到容斥原理。

　　先假設所有的集合的gcd是i的倍數都符合條件然後計算答案(給定數集A中所有是i的倍數的數組成的集合任選一個子集)，然後再減去f(2i), f(3i)，...

　　現在要面臨兩個問題

1. 第一次求值如何處理？
   首先把式子寫出來，設這個集合的大小為n，那麽有

   因為

   對兩邊同時進行求導得到

   再帶入x = 1得到

2. 設為是i的倍數的數的個數，如何快速求出?

   根據定義式有

   顯然超時。雖然這是暴力，但是不夠優美。
   設表示，集合A中恰好為i的數有多少個。

   然後就可以得到總時間復雜度為O(mlog₂m)的暴力:

　　最後求求和就完事了。

Code

 1 /**
 2  * Codeforces
 3  * Problem#839D
 4  * Accepted
 5  * Time: 171ms
 6  * Memory: 15400k
 7  */
 8 #include <bits/stdc++.h>
 9 using namespace std;
10 
11 const int lim = 1e6 + 1;
12 const int moder = 1e9 + 7;
13 
14 int n;
15 int *a;
16 int *pow2;
17 int cnt[lim], counter[lim];
18 int f[lim];
19 int res = 0;
20 
21 inline void init() {
22     scanf("%d", &n);
23     a = new int[(n + 1)];
24     pow2 = new int[(n + 1)];
25     pow2[0] = 1;
26     for(int i = 1; i <= n; i++) {
27         scanf("%d", a + i);
28         counter[a[i]]++;
29         pow2[i] = (pow2[i - 1] << 1) % moder;
30     }
31 }
32 
33 inline void solve() {
34     for(int i = 1; i < lim; i++)
35         for(int j = i; j < lim; j += i)
36             cnt[i] += counter[j];
37     
38     for(int i = lim - 1; i > 1; i--) {
39         if(!cnt[i])    continue; 
40         f[i] = (cnt[i] * 1LL * pow2[cnt[i] - 1]) % moder;
41         for(int j = i << 1; j < lim; j += i)
42             f[i] = (f[i] - f[j]) % moder;
43         if(f[i] < 0)    f[i] += moder;
44         res = (res + (f[i] * 1LL * i) % moder) % moder;
45     }
46     
47     printf("%d\n", res);
48 }
49 
50 int main() {
51     init();
52     solve();
53     return 0;
54 }

Codeforces 839D Winter is here - 暴力 - 容斥原理