【LOJ】#2047. 「CQOI2016」伪光滑数

题解

可持久化可并堆

用\(f[i,j]\)表示最大的质数标号为i，然后有j个质数乘起来

用\(g[i,j]\)表示\(\sum_{k = 1}^{i}f[k,j]\)

转移是

\(f[i,j] = \sum_{k= 1}^{j} g[i - 1,j - k] * p_{i}^{k}\)

\(g[i,j] += f[i,j]\)

这就要资瓷两个集合的合并了

可是集合显然非常大，我们可以用可持久化来帮助完成这件事，就完成了

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 1000005
#define pb push_back
#define mp make_pair
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int64 N;
int K,prime[105],tot;
bool nonprime[205];
int64 pw[105];
struct node {
    node *lc,*rc;
    int64 mul,val;
    int dis;
}pool[17000000],*tail = pool;
struct set_node {
    int64 val;node *rt;
    set_node(){}
    set_node(int64 _v,node *r) {
        val = _v;rt = r;
    }
    friend bool operator < (const set_node &a,const set_node &b) {
        return a.val > b.val;
    }
    friend bool operator == (const set_node &a,const set_node &b) {
        return a.val == b.val && a.rt == b.rt;
    }
};
node *g[105],*f[105];
set<set_node > S;
node *Newnode(int64 v) {
    node *res = tail++;
    res->mul = 1;res->val = v;
    res->lc = res->rc = NULL;
    res->dis = 0;
    return res;
}
int getdist(node *p) {
    return p ? p->dis : -1;
}
node* addlazy(node *u,int64 val) {
    if(!u) return NULL;
    node *res = tail++;
    *res = *u;
    res->val *= val;
    res->mul *= val;
    return res;
}
void push_down(node *&u) {
    if(u->mul != 1) {
        u->lc = addlazy(u->lc,u->mul);
        u->rc = addlazy(u->rc,u->mul);
        u->mul = 1;
    }
}
node *Merge(node *A,node *B) {
    if(!A) return B;
    if(!B) return A;
    if(A->val < B->val) swap(A,B);
    push_down(A);
    node *res = tail++;
    *res = *A;
    res->rc = Merge(A->rc,B);
    if(getdist(res->rc) > getdist(res->lc)) swap(res->lc,res->rc);
    res->dis = getdist(res->rc) + 1;
    return res;
}
void Solve() {
    read(N);read(K);
    for(int i = 2 ; i <= 128 ; ++i) {
        if(!nonprime[i]) {
            prime[++tot] = i;
            for(int j = 2 ; j <= 128 / i ; ++j) {
                nonprime[i * j] = 1;
            }
        }
    }
    for(int i = 0 ; i <= 100 ; ++i) g[i] = NULL;
    g[0] = Newnode(1);
    for(int i = 1 ; i <= tot ; ++i) {
        int cnt = 0;int64 t = N;
        while(t >= prime[i]) {t /= prime[i];++cnt;}
        pw[0] = 1;
        for(int k = 1 ; k <= cnt ; ++k) pw[k] = pw[k - 1] * prime[i];
        for(int k = 1 ; k <= cnt ; ++k) {
            f[k] = NULL;
            for(int h = 1 ; h <= k ; ++h) {
                f[k] = Merge(f[k],addlazy(g[k - h],pw[h]));
            }
        }
        for(int k = 1 ; k <= cnt ; ++k) g[k] = Merge(g[k],f[k]);
    }
    node *rt = NULL;
    for(int i = 1 ; i <= 100 ; ++i) rt = Merge(rt,g[i]);
    S.insert(set_node(rt->val,rt));
    for(int i = 1 ; i < K ; ++i) {
        set_node p = *S.begin();
        S.erase(S.begin());
        push_down(p.rt);
        if(p.rt->lc) S.insert(set_node(p.rt->lc->val,p.rt->lc));
        if(p.rt->rc) S.insert(set_node(p.rt->rc->val,p.rt->rc));
        while(S.size() > K - i) S.erase(--S.end());
    }
    out((*S.begin()).val);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}