Codeforces-258C：Little Elephant and LCM（数论）

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概述

The Little Elephant loves the LCM (least common multiple) operation of a non-empty set of positive integers. The result of the LCM operation of k positive integers x1, x2, ..., xk is the minimum positive integer that is divisible by each of numbers xi.

Let's assume that there is a sequence of integers b1, b2, ..., bn. Let's denote their LCMs as lcm(b1, b2, ..., bn) and the maximum of them as max(b1, b2, ..., bn). The Little Elephant considers a sequence b good, if lcm(b1, b2, ..., bn) = max(b1, b2, ..., bn).

The Little Elephant has a sequence of integers a1, a2, ..., an. Help him find the number of good sequences of integers b1, b2, ..., bn, such that for all i (1 ≤ i ≤ n) the following condition fulfills: 1 ≤ bi ≤ ai. As the answer can be rather large, print the remainder from dividing it by 1000000007 (109 + 7).

Input

The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of integers in the sequence a. The second line contains nspace-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105) — sequence a.

Output

In the single line print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

   Examples 
 
     input 
   
4
1 4 3 2

     output 
   
15

     input 
   
2
6 3

     output 
   
13

思路：将数列a排序（改变a中元素排列顺序不改变答案），枚举最大值，那么b数列中其他的值一定是那个最大值的因子。那么找出其所有因子并排序。二分找出包含这个因子的a序列的区间。

如：a=[1, 2 ,3 ,4 ] 4

枚举最大值 2，2，2

首先是4，a数列中各数包含4的因子分别为 1，1，1，1 ，那么最大值为4时的b数列的方案数为1*2*2*（4-3）=4种。因为要满足最大值为4，所以最后只能取4.

3，3

其次是3，a数列中各数包含3的因子分别为 1，1 ，1，1 ，那么最大值为3时的b数列的方案数为1*1*（2*2-1）=3种。因为要满足最大值为3，所以最后有3种取法。

以此类推。

#include<bits/stdc++.h>
typedef __int64 ll;
using namespace std;
const int MOD=1e9+7;
const int MAX_N = 1e6;
int all=0,p[MAX_N],m;
int pr[MAX_N+100],a[MAX_N];
bool isp[MAX_N+10];
void init()
{
	all = 0;
	memset(isp,0,sizeof isp);
	isp[1]=1;
	for(int i=2;i<=MAX_N;i++)
	{
		if(!isp[i])pr[all++] = i;
		for(int j=0;j<all;j++)
		{
			ll t = (ll)1*pr[j]*i ;
			if(t<=MAX_N)
			{
				isp[t] = true;
				if(i%pr[j]==0)break;
			}
			else break;
		}
	}
}
void dive(int x)
{
    m=0;
    for(int i=1;i*i<=x;i++)
    {
        if(x%i==0)
        {
            p[m++]=i;
            if(x/i!=i)p[m++]=x/i;
        }
    }
    sort(p+1,p+m);
}
ll POW(ll x,ll y)
{
    ll res=1;
    while(y)
    {
        if(y&1)res=res*x%MOD;
        x=(x*x)%MOD;
        y/=2;
    }
    return res%MOD;
}
int main()
{
    init();
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&a[i]);
    sort(a,a+n);
    ll ans=0;
    for(int i=a[n-1];i>=1;i--)
    {
        dive(i);
        ll st=0,now=1;
        for(int j=1;j<m;j++)
        {
            ll en=lower_bound(a,a+n,p[j])-a;
            now=now*POW(j,en-st)%MOD;
            st=en;
        }
        now=now*(((POW(m,n-st)-POW(m-1,n-st))%MOD+MOD)%MOD)%MOD;
        ans=(ans+now)%MOD;
    }
    cout<<ans<<endl;
}