3.1.3 Humble Numbers

Humble Numbers

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

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

/*
ID: makeeca1
PROG: humble
LANG: C++
*/
#include <cstdio>
#include<climits>
using namespace std;
#define MAX 200
int k,n,a[MAX],ans[100001],dex[MAX],count,min;
int main(){
    freopen("humble.in","r",stdin);
    freopen("humble.out","w",stdout);
    scanf("%d%d",&k,&n);
    for (int i=0;i<k;i++) scanf("%d",&a[i]);
    count=1;
    ans[0]=1;
    while (count<n+1){
        min=INT_MAX;
        for (int i=0;i<k;i++){
            while ((long long)a[i]*ans[dex[i]]<=ans[count-1])dex[i]++;
            if ((long long )a[i]*ans[dex[i]]<min)min=a[i]*ans[dex[i]];
        }
        ans[count++]=min;
    }
    printf("%d
",ans[n]);
    return 0;
}
原文地址:https://www.cnblogs.com/makeecat/p/3289222.html