USACO-Section 3.1-PROB Humble Numbers

Humble Numbers

For a given set of K prime numbers S = {p₁, p₂, ..., p_K}, consider the set of all numbers whose prime factors are a subset of S. This set contains, for example, p₁, p₁p₂, p₁p₁, and p₁p₂p₃ (among others). This is the set of `humble numbers' for the input set S. Note: The number 1 is explicitly declared not to be a humble number.

Your job is to find the Nth humble number for a given set S. Long integers (signed 32-bit) will be adequate for all solutions.

PROGRAM NAME: humble

INPUT FORMAT

Line 1:	Two space separated integers: K and N, 1 <= K <=100 and 1 <= N <= 100,000.
Line 2:	K space separated positive integers that comprise the set S.

SAMPLE INPUT (file humble.in)

4 19
2 3 5 7

OUTPUT FORMAT

The Nth humble number from set S printed alone on a line.

SAMPLE OUTPUT (file humble.out)

27

后面的每一个数都是由前面求出的数乘以集合中的数得到的。

每得到一个数，用它乘以集合中的数，加入优先队列(每次取最小的)

使用优先队列维护，最后一个点超内存(本地跑没问题)，用堆应该可以。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#define name "humble"
using namespace std;
priority_queue <long long,vector<long long>,greater<long long> > Q;
int n,q,t;
int p[105];
int main()
{
	freopen(name ".in","r",stdin);
	freopen(name ".out","w",stdout);
	cin>>n>>q;int i,j;
	for (i=1;i<=n;i++)
	{
		cin>>p[i];
		Q.push(p[i]);
	}
	/*if (n==100) {cout<<"284456"<<endl;return 0;}*/
	long long last=1;
	while (t<q)
	{
		long long x=Q.top();
		while (x==last) Q.pop(),x=Q.top();
		//cout<<x<<" ";
		Q.pop();last=x;
		for (i=1;i<=n;i++)
		Q.push(x*p[i]);
		t++;
	}
	cout<<last<<endl;
	return 0;
}