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

C. Little Elephant and LCMtime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe 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 positiveintegers x1, x2, ..., xk isthe 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) andthe 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 followingcondition fulfills: 1 ≤ bi ≤ ai.As the answer can be rather large, print the remainder from dividing it by 1000000007 (109 + 7).InputThe first line contains a single positive integer n (1 ≤ n ≤ 105) —the number of integers in the sequence a. The second line contains nspace-separatedintegers a1, a2, ..., an (1 ≤ ai ≤ 105) —sequence a.OutputIn 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 <iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<algorithm>
#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;i<=a
;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;
}

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：