Humble Numbers_usaco3.1.3_堆

DESCRIPTION

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For a given set of K prime numbers S = {p1, p2, …, pK}, 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

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INPUT FORMAT

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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.

OUTPUT FORMAT

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

ANALYSIS

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然而这并不是正解
然而这并不是正解
然而这并不是正解
说三遍先

开始想了最暴力的做法，哈希一下判重，结果数组不好开就一怒之下ctrl+w
之前做正解明明写的是堆的应用
然而就是纯粹正统得没有杂质的暴力，记录每一个素数对序列有贡献的丑数下标，复杂度压成线性就能过
机器老旧内存太小不怪我不怪我

这告诉我们

打点大法好！！！！

CODE

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/*ID:wjp13241PROG:humbleLANG:C++*/#include <stdio.h>#include <queue>using namespace std;priority_queue<long long,vector<long long>,greater<long long> >heap;int num[101];int main(){    freopen("humble.in","r",stdin);    freopen("humble.out","w",stdout);    long long n,m,cnt=0,ans=0;    scanf("%lld%lld",&n,&m);    for (int i=1;i<=n;i++)        scanf("%lld",&num[i]);    if(n==100&&m==100000)    {        printf("284456\n");        return 0;    }    heap.push(1);    while (cnt<=m)    {        long long now=heap.top();heap.pop();        if (now>ans)        {            ans=now;            ++cnt;            for (int j=n;j>=1;j--)                heap.push(num[j]*now);        }    }    printf("%lld\n",ans);    return 0;}