HackerEarth Benny and the Universe problem solution

In this HackerEarth Benny and the Universe problem solution Hope you still have not forgotten about pig Benny. She is asking for help again. This time you have to answer some queries that Benny will give you.

All actions happen in the Universe where each planet has its own id starting from 0. There are infinite amount of planets.

To travel between different planets there are some special spacecrafts. Each spacecraft is characterized by the number di.

If the id of the planet you are currently staying on is idj and you decided you use spacecraft with di you will go to planet with id equal to idj + di.

Benny’s parents live nearby the planet with id equal to 0, so you may assume that there exists some spacecraft for which di is not more than 104.

You have to answer Q queries. Each query consists of one single integer x. You have to answer whether it’s possible to go from planet with id equal to 0 to planet with id equal to x.

HackerEarth Benny and the Universe problem solution.

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <utility>
#include <memory.h>
#include <cassert>
#include <iterator>
#include <bitset>
#include <iomanip>
#include <complex>
#include <queue>
#include <ctime>
#include <deque>
#include <stack>
#include <set>
#include <map>
 
using namespace std;
 
#define pb push_back
#define mp make_pair
#define F first
#define S second

const int N = 1010;
const long long INF = (long long)2e18;

int n, q;
int a[N];
long long f[10 * N];

int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    int minAi = a[1];
    for (int i = 1; i <= n; i++) {
        minAi = min(minAi, a[i]);
    }
    for (int i = 0; i <= minAi; i++) {
        f[i] = INF;
    }
    priority_queue< pair<long long, int> > pq;
    f[0] = 0LL;
    pq.push(mp(0, 0));
    while (!pq.empty()) {
        long long dist = -pq.top().F;
        int ver = pq.top().S;
        pq.pop();
        if (f[ver] < dist) {
            continue;
        }
        for (int i = 1; i <= n; i++) {
            int nxt = (ver + a[i]) % minAi;
            if (f[nxt] > f[ver] + 1LL * a[i]) {
                f[nxt] = f[ver] + 1LL * a[i];
                pq.push(mp(-f[nxt], nxt));
            }
        }
    }
    bool pp, mm;
    pp = mm = false;
    while (q--) {
        int ver;
        scanf("%d", &ver);
        int real = ver;
        ver %= minAi;
        if (f[ver] <= 1LL * real) {
            puts("YES");
            pp = true;
        } else {
            puts("NO");
            mm = true;
        }
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>

using namespace std;

const int N = 10000;

int n, tests;
int d[N];
set<pair<int, int> > S;
set<pair<int, int> >::iterator it;
int dist[N];

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> tests;

    for (int i = 0; i < n; i++)
    {
        cin >> d[i];
    }

    sort(d, d + n);

    for (int i = 0; i < d[0]; i++)
    {
        dist[i] = 2e9+1e6;
    }

    dist[0] = 0;

    for (int i = 0; i < d[0]; i++)
    {
        S.insert(make_pair(dist[i], i));
    }

    while (S.size())
    {
        it = S.begin();
        int v = (*it).second;
        S.erase(it);
        if (dist[v]>1e9 + 1e8)
            continue;

        for (int i = 0; i < n; i++)
        {
            int new_val = dist[v] + d[i];
            int rem = new_val%d[0];

            if (dist[rem]>new_val)
            {
                S.erase(make_pair(dist[rem], rem));
                dist[rem] = new_val;
                S.insert(make_pair(dist[rem], rem));
            }
        }
    }

    for (; tests; --tests)
    {
        int val;
        cin >> val;
        long long R = val%d[0];
        if (dist[R] <= val)
        {
            cout << "YESn";
        }
        else
        {
            cout << "NOn";
        }
    }

    return 0;
}

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