Codeforces Round #157 (Div. 1) C. Little Elephant and LCM （数学、dp）_PHP教程

Codeforces Round #157 (Div. 1) C. Little Elephant and LCM （数学、dp）

C. Little Elephant and LCM time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output

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).

Sample test(s)
input

4
1 4 3 2

output

15

input

2
6 3

output

13

题意：

给你一个a序列，找出一个b序列，1?≤?bi?≤?ai，使得max(bi)=lcm(bi)，问这样的bi序列有多少个。

思路：

先对a排序，枚举i=max(bi)，对i因式分解，那么大于等于i的部分很好处理，直接pow_mod()相减，小于i的部分就任意取一个约束就够了。

代码：

#include 
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
#define maxn 100005
#define mod 1000000007
typedef long long ll;
using namespace std;

int n;
int a[maxn];

ll pow_mod(ll x,ll n)
{
    ll res = 1;
    while(n)
    {
        if(n&1) res = res * x %mod;
        x = x * x %mod;
        n >>= 1;
    }
    return res;
}
void solve()
{
    int i,j;
    ll ans=0,res;
    sort(a+1,a+n+1);
    for(i=1;ifac;
        for(j=1;j*j
						

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