CodeforcesRound #157 (Div.1) C. LittleElephantandLCM (Mathematics, dp) _ PHP Tutorial

CodeforcesRound #157 (Div.1) C. LittleElephantandLCM (Mathematics, dp ). CodeforcesRound #157 (Div.1) C. LittleElephantandLCM (Mathematics, dp) C. parse Codeforces Round #157 (Div. 1) C. Little Elephant and LCM (Mathematics, dp)
C. Little Elephant and LCMtime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

The Little Elephant loves the LCM (least common multiple) operation of a non-empty set of positive integers. The result of the LCM operationKPositive integersX1 ,?X2 ,?...,?XKIs the minimum positive integer that is pisible by each of numbersXI.

Let's assume that there is a sequence of integersB1 ,?B2 ,?...,?BN. Let's denote their LCMsLcm(B1 ,?B2 ,?...,?BN) And the maximum of themMax(B1 ,?B2 ,?...,?BN). The Little Elephant considers a sequenceBGood, ifLcm(B1 ,?B2 ,?...,?BN)? =?Max(B1 ,?B2 ,?...,?BN).

The Little Elephant has a sequence of integersA1 ,?A2 ,?...,?AN. Help him find the number of good sequences of integersB1 ,?B2 ,?...,?BN, Such that for allI(1? ≤?I? ≤?N) The following condition fulfills: 1? ≤?BI? ≤?AI. As the answer can be rather large, print the remainder from piding it by 1000000007 (109? +? 7 ).

Input

The first line contains a single positive integerN(1? ≤?N? ≤? 105)-the number of integers in the sequenceA. The second line containsNSpace-separated integersA1 ,?A2 ,?...,?AN(1? ≤?AI? ≤? 105)-sequenceA.

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

Question:

Let's give you a sequence a and find a sequence B, 1? ≤? Bi? ≤? Ai, so that max (bi) = lcm (bi), ask how many such bi sequences there are.

Ideas:

Sort a first, enumerate I = max (bi), and break down the I-type decomposition. the part greater than or equal to I is well processed, and the pow_mod () is directly subtracted, if the number is smaller than I, any constraint is enough.

Code:

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# Include # define INF 0x3f3f3f # define maxn 100005 # define mod 1000000007 typedef 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; I <= a [n]; I ++) // enumerated answer {vector
       
         Fac; for (j = 1; j * j <= I; j ++) // factor {if (I % j = 0) {fac. push_back (j); if (j * j! = I) fac. push_back (I/j);} sort (fac. begin (), fac. end (); int pos, pre = 1; res = 1; for (j = 1; j
        
       
      
     
    
   
  
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