Codeforces Round #157 (Div. 1) C. Little Elephant and LCM （数学、dp）_html/css_WEB-ITnose

C. Little Elephant and LCM

time limit per test

memory limit per test

input

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.

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Output

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

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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：

<pre name="code" class="n">#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;i<=a[n];i++) // 枚举答案    {        vector<int>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<fac.size();j++)        {            pos=lower_bound(a+1,a+n+1,fac[j])-a;            res*=pow_mod(j,pos-pre);            res%=mod;            pre=pos;        }        ans+=res*((pow_mod(j,n-pre+1)-pow_mod(j-1,n-pre+1)+mod)%mod);        ans%=mod;    }    printf("%I64d\n",ans);}int main(){    int i,j;    while(~scanf("%d",&n))    {        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        solve();    }    return 0;}