kotori和素因子 (质因数分解+DFS）

https://ac.nowcoder.com/acm/contest/940/E

$题解：先对每个数分解质因数，然后就是裸DFS。$

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include 
using namespace std;
#define inf 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x)  x&-x;  
#define debugint(name,x) printf("%s: %d\n",name,x);
#define debugstring(name,x) printf("%s: %s\n",name,x);
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-6;
const int maxn = 1e5+5;
const int mod = 1e9+7;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,a[1005];
vector<int>num[1005];
int vis[1005],ok = 0,ans = inf;
void dfs(int id,int cnt,int res){
    if(cnt == n){
        ans = min(ans,res);
        ok = 1;
        return ;
    }
    int sz = num[a[id]].size();
    for(int i = 0; i < sz; i++){
        int tmp = num[a[id]][i];
        if(vis[tmp]) continue;
        vis[tmp] = 1;
        dfs(id+1,cnt+1,res+tmp);
        vis[tmp] = 0;
    }
}
int main(){
    cin>>n;
    int flag = 0;
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        int cp = a[i];
        for(int j = 2; j <= sqrt(a[i]); j++){
            if(cp%j==0){
                num[a[i]].push_back(j);
                while(cp%j==0){
                    cp /= j;
                }
            }
        }
        if(cp > 1) num[a[i]].push_back(cp);
    }
    dfs(1,0,0);
    if(ok != 1) puts("-1");
    else printf("%d\n",ans);
}