Codeforces 839 D. Winter is here (莫比乌斯反演)

Description

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-th soldier’s strength be ai. For some k he calls i1, i2, ..., ik a clan if i1 < i2 < i3 < ... < ik and gcd(ai1, ai2, ..., aik) > 1 . He calls the strength of that clan k·gcd(ai1, ai2, ..., aik). Then he defines the strength of his army by the sum of strengths of all possible clans.

Your task is to find the strength of his army. As the number may be very large, you have to print it modulo 1000000007 (10^9 + 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.

 

Input

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.

 

Output

Print one integer — the strength of John Snow's army modulo 1000000007 (10^9 + 7).

 

Example input

4
2 3 4 6

 

Example output

39

 

题意

给出一个序列,求其中所有 $\gcd$ 大于 1 的子序列乘以序列长度之和。

 

思路

题意转化为公式即 $\sum k×\gcd(a_1,a_2,...,a_k)$ (其中 $\gcd$ 大于 1)

等价于 $\sum_{d=2}^{max}d×\sum k×[\gcd(a_1,a_2,...,a_k)=d]$

我们设 $f(x)=\sum k×[gcd(a_1,a_2,...,a_k)=x]$

设 $F(x)=\sum_{x|d}f(d)$ ,它也代表我们从整个数列所有 $x$ 的倍数中挑选 $k$ 个组成的情况,因此我们设 $v$ 为数列中 $x$ 倍数的个数,则结果为 $1×C_v^1+2×C_v^2+...+v×C_v^v$ ,即 $v×2^{v-1}$ 。

反演以后得到 $f(x)=\sum_{x|d}μ(\frac{d}{x})F(d)$ ,最终结果 $ans=\sum_{d=2}^{max}d×f(d)$ 。


考虑枚举每一个 $\gcd​$ ,找出序列中所有包含该因子的数有 $V_{gcd}​$ 个,则此时该数对结果的贡献为 $V_{gcd}×2^{V_{gcd}-1}-S​$ ,其中 $S​$ 为其所有倍数所带来的贡献。

最终结果 $ans=\sum_{d=2}^{max}d×(V_{d}×2^{V_{d}-1}-S)$

 

AC 代码

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int mod = 1e9+7;
const int maxn = 1e6+10;
LL sk[maxn],b[maxn],z[maxn];

void solve()
{
    LL ans=0;
    for (int i=2; i<=maxn; i++)
    {
        LL v=0;
        for (int j=i; j<=maxn; j+=i)
            v+=sk[j];
        if (!v) continue;
        ans += (v * z[i]) % mod * b[v-1] % mod;
    }
    cout << (ans+mod)%mod << endl;
}

void init()
{
    b[0]=1;
    for (int i=1; i<=maxn; i++)
        b[i]=b[i-1]*2%mod,z[i]=i;
    for (int i=2; i<=maxn; i++)
        for (int j=i+i; j<=maxn; j+=i)
            z[j]-= z[i];
}

int main()
{
    init();
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        memset(sk,0,sizeof(sk));
        for (int i=1,x; i<=n; i++)
        {
            cin>>x;
            sk[x]++;
        }
        solve();
    }
    return 0;
}