題意：

若$a_1+a_2+\cdots+a_h=n$(任意h<=n),求$lcm(a_i)$的種類數

思路：

設$lcm(a_i)=x$，

由唯一分解定理，$x=p_1^{m_1}+p_2^{m_2}+\cdots+p_{tot}^{m_{tot}}$

設$b_i=p_i^{m_i}$，

則能組成x的和最小的數為$\sum p_i^{m_i}$

所以只要$\sum p_i^{m_i}\leq n$即可，

其中小於的時候，剩餘補1即可

dp[i][j]表示選了前i個素數，他們的和為j時的方法數

程式碼：

#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-6;
const int mod = 1e9+7;
const int maxn = 2e3+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n, tot;
int prime[1000 + 10];
int vis[1000 + 10];
ll ans, dp[1000 +10][1000 + 10];
int main(){
    scanf("%d", &n);
    tot = 0;
    for(int i = 2; i <= 1000; i++){
        if(!vis[i])prime[++tot] = i;
        for(int j = 1; j <= tot && i *prime[j] <= 1000; j++){
            vis[i*prime[j]] = 1;
            if(i%prime[j]==0)break;
        }
    }
    dp[0][0] = 1;
    for(int i = 1; i <= tot; i++){
        for(int j = 0; j <= n; j++)dp[i][j] = dp[i-1][j];
        for(int j = prime[i]; j <= n; j *= prime[i]){
            for(int k = 0; k + j <= n; k++){
                dp[i][k+j] += dp[i-1][k];
            }
        }
    }
    ans = 0;
    for(int i = 0; i <= n; i++)ans+=dp[tot][i];
    printf("%lld", ans);
    return 0;
}