839D(容斥 || 莫比乌斯反演)

最新推荐文章于 2019-11-06 20:09:08 发布

creatorx

最新推荐文章于 2019-11-06 20:09:08 发布

阅读量457

点赞数 1

分类专栏：莫比乌斯反演

本文链接：https://blog.youkuaiyun.com/creatorx/article/details/77164886

版权

莫比乌斯反演专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一个算法问题，通过莫比乌斯反演计算军队的总体强度。士兵们的强度由其战斗力决定，而军队的整体强度则是所有可能组合的士兵群体（称为氏族）的强度之和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 (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.

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 (109 + 7).

   Examples 
 
     input 
   
3
3 3 1

     output 
   
12

     input 
   
4
2 3 4 6

     output 
   
39

Note

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

这种gcd问题，肯定是要用到容斥或者莫比乌斯反演，这里我用的是莫比乌斯反演。

假设答案为ans.则根据题意可得：

ans = Σd * Σk * (gcd(ai1, ai2,,,,,aik) == d)

我们枚举d就行。

我们令f(n) = Σk * (gcd(ai1, ai2,,,,,aik) == n)

g(n) = Σf(d) (n|d) = num * 2^(num -1)(num为数组中能够整除n的数的个数）

然后我们可以得到：

f(n) = Σmu(d / n)g(d) (n|d)

然后代回原公式计算就行

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod = 1e9 + 7;
const ll maxn = 1000000 + 10;
ll n;
ll ans;
ll times[maxn];//原数组中每个元素出现的次数
ll term[maxn];
ll g[maxn];
ll f[maxn];
ll prime[maxn];
ll mu[maxn];//莫比乌斯函数
bool valid[maxn];
ll quick_pow(ll x, ll i)
{
    ll result = 1;
    x %= mod;
    while(i)
    {
        if(i&1) result = (result * x) % mod;
        i >>= 1;
        x = (x * x) % mod;
    }
    return result;
}
void initMu()
{
    mu[1] = 1;
    memset(valid, true, sizeof(valid));
    ll cnt = 0;
    for(ll i = 2; i < maxn; i++)
    {
        if(valid[i])
        {
            prime[++cnt] = i;
            mu[i] = -1;
        }
        for(ll j = 1; j <= cnt && i * prime[j] < maxn; j++)
        {
            valid[i * prime[j]] = false;
            if(i % prime[j] == 0)
            {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
}
void initCase()
{
    memset(times, 0, sizeof(times));
    memset(term, 0, sizeof(term));
    ans = 0;
}
void initG()
{

    for(ll i = 2; i < maxn; i++)
    {
        for(ll j = i; j < maxn; j += i)
        {
            times[i] += term[j];
        }
    }

    for(ll i = 2; i < maxn; i++)
    {
        ll res = times[i];
        if(res)
        g[i] = ((res % mod) * quick_pow(2, res - 1)) % mod;
        else g[i] = 0;
    }
}
void initF()
{
    for(ll i = 2; i < maxn; i++)
    {
        f[i] = 0;
        for(ll j = i; j < maxn; j += i)
        {
            f[i] = (f[i] + mu[j / i] * g[j]) % mod;
        }
    }
}
void initAns()
{
    for(ll i = 2; i < maxn; i++)
    {
        ans = (ans + i * f[i] % mod) % mod;
    }
}
int main()
{
    initMu();
    while(~scanf("%lld", &n))
    {
        ll value;
        initCase();
        for(ll i = 1; i <= n; i++)
        {
            scanf("%lld", &value);
            term[value]++;
        }
        initG();
        initF();
        initAns();
        printf("%I64d\n", ans % mod);
    }
    return 0;
}