Codeforces 986F Oppa Funcan Style Remastered

Posted 2021-11-29 heltion

986F Oppa Funcan Style Remastered

题意:给定\(n,k\),是否存在\(p\in S_n,p\ne 1,p^k=1.\)

题解:\(k=\displaystyle\prod_{i=1}^mp_i^{r_i},\)问题等价于\(n=\displaystyle\sum_{i=1}^mp_ix_i\)是否有非负整数解.定义\(d(r)\)为最小使得\(d=\displaystyle\sum_{i=1}^mp_ix_i\equiv r(\text{mod }p_1)\)有非负整数解的\(d\),那么有解等价于\(n\geq d(n\mod p_1).m=2\)时可以用扩展Euclid算法求解.\(m\geq 3\)时可以用Dijkstra算法求解.

省略了扩展Euclid模板部分.

#include<bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int maxt = 12000;
constexpr int maxk = 3.2e7;
constexpr int maxn = 1.2e5;
//https://cp-algorithms.com/algebra/extended-euclid-algorithm.html
struct query{
    LL n, k, id;
    bool operator < (const query &b){
        return k < b.k;
    }
}q[maxt];
bool vd[maxn], vp[maxk], ans[maxt];
LL d[maxn];
vector<LL> v, p;
void factorize(LL k){
    p.clear();
    for(int i = 0; k > 1 and i < (int)v.size(); i += 1)
        if(k % v[i] == 0){
            p.push_back(v[i]);
            while(k % v[i] == 0) k /= v[i];
        }
    if(k > 1) p.push_back(k);
    if(p.size() > 2){
        fill(d + 1, d + p[0], 1e18);
        fill(vd, vd + p[0], false);
        priority_queue<pair<LL, int>, vector<pair<LL, int>>, greater<pair<LL, int>>> q;
        q.push(make_pair(d[0], 0));
        while(not q.empty()){
            int u = q.top().second;
            q.pop();
            if(vd[u]) continue;
            vd[u] = true;
            for(const int &w : p){
                int v = (u + w) % p[0];
                if(d[v] > d[u] + w) q.push(make_pair(d[v] = d[u] + w, v));
            }
        }
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    for(int i = 2; i < maxk; i += 1){
        if(not vp[i]) v.push_back(i);
        for(LL p : v){
            if(p * i >= maxk) break;
            vp[p * i] = true;
            if(i % p == 0) break;
        }
    }
    int t;
    cin >> t;
    for(int i = 0; i < t; i += 1){
        cin >> q[i].n >> q[i].k;
        q[i].id = i;
    }
    sort(q, q + t);
    for(int i = 0; i < t; i += 1){
        if(not i or q[i].k != q[i - 1].k) factorize(q[i].k);
        if(p.size() == 0) ans[q[i].id] = false;
        else if(p.size() == 1) ans[q[i].id] = q[i].n % p[0] == 0;
        else if(p.size() == 2){
            LL x, y;
            gcd(p[0], p[1], x, y);
            y = (y % p[0] + p[0]) % p[0];
            LL dr = y * (q[i].n % p[0]) % p[0] * p[1];
            ans[q[i].id] = dr <= q[i].n;
        }
        else ans[q[i].id] = d[q[i].n % p[0]] <= q[i].n;
    }
    for(int i = 0; i < t; i += 1)
        cout << (ans[i] ? "YES\n" : "NO\n");
    return 0;
}