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.
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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
41 4 3 2
output
15
input
26 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 <iostream>#include<cstring>#include<cstdio>#include<string>#include<vector>#include<algorithm>#define INF 0x3f3f3f3f#define maxn 100005#define mod 1000000007typedef 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 </algorithm></vector></string></cstdio></cstring></iostream>
