Codeforces Round #428 (Div. 2) - D Winter is here

最新推荐文章于 2019-03-28 16:42:32 发布

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本文介绍了一个算法问题，模拟《权力的游戏》中约翰·雪诺评估其军队实力的方式。任务是通过计算所有士兵组合的加权总和来确定军队的整体实力，其中每个组合的权重由士兵数量与最大公约数的乘积决定。

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

题意：

找出所有gcd>1的组合，他们的价值为gcd*个数。求总价值。

POINT：

遍历所有GCD=b，找出以b为gcd的数的个数x。那么它的价值为：b*(0*C(x,0)+1*C(x,1)+2*C(x,2)+……+x*C(x,x)).

可化为x*2^(x-1).怎么化的下面有两张图。

然后用容斥。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define LL long long
const LL N = 1000000;
const LL p = 1e9+7;
LL num[N+1];
LL pos[N+1];
LL f[N+1];
void init()
{
    memset(f,0,sizeof f);
    pos[0]=1;
    for(LL i=1;i<=N;i++)
    {
        pos[i]=pos[i-1]*2%p;
    }
}
int main()
{
    init();
    LL n;
    scanf("%lld",&n);
    for(LL i=1;i<=n;i++)
    {
        LL a;scanf("%lld",&a);
        num[a]++;
    }
    LL ans=0;
    for(LL i=N;i>=2;i--)
    {
        LL x=0;
        for(LL j=i;j<=N;j+=i)
        {
            x+=num[j];
        }
        if(x==0) continue;
        f[i]=x*pos[x-1]%p;
        for(LL j=i*2;j<=N;j+=i)
        {
            f[i]-=f[j];
            f[i]=(f[i]+p)%p;
        }
        ans=(ans+i*f[i])%p;
    }
    printf("%lld\n",ans);
}