题意：
若$a_1+a_2+\cdots+a_h=n$(任意h<=n),求$lcm(a_i)$的种类数
思路：
设$lcm(a_i)=x$，
由唯一分解定理，$x=p_1^{m_1}+p_2^{m_2}+\cdots+p_{tot}^{m_{tot}}$
设$b_i=p_i^{m_i}$，
则能组成x的和最小的数为$\sum p_i^{m_i}$
所以只要$\sum p_i^{m_i}\leq n$即可，
其中小于的时候，剩余补1即可
dp[i][j]表示选了前i个素数，他们的和为j时的方法数
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional> #define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) using namespace std; typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL; const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e3+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0); int n, tot;
int prime[ + ];
int vis[ + ];
ll ans, dp[ +][ + ];
int main(){
scanf("%d", &n);
tot = ;
for(int i = ; i <= ; i++){
if(!vis[i])prime[++tot] = i;
for(int j = ; j <= tot && i *prime[j] <= ; j++){
vis[i*prime[j]] = ;
if(i%prime[j]==)break;
}
}
dp[][] = ;
for(int i = ; i <= tot; i++){
for(int j = ; j <= n; j++)dp[i][j] = dp[i-][j];
for(int j = prime[i]; j <= n; j *= prime[i]){
for(int k = ; k + j <= n; k++){
dp[i][k+j] += dp[i-][k];
}
}
}
ans = ;
for(int i = ; i <= n; i++)ans+=dp[tot][i];
printf("%lld", ans);
return ;
}