This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

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.

### 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 compose 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
``````

Java

C++

Pascal

## ANALYSIS

Russ Cox

We compute the first n humble numbers in the “hum” array. For simplicity of implementation, we treat 1 as a humble number, and adjust accordingly.

Once we have the first k humble numbers and want to compute the k+1st, we do the following:

``````for each prime p
find the minimum humble number h
such that h * p is bigger than the last humble number.

take the smallest h * p found: that's the next humble number.
``````

To speed up the search, we keep an index “pindex” of what h is for each prime, and start there rather than at the beginning of the list.

``````#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>

#define MAXPRIME 100
#define MAXN 100000

long hum[MAXN+1];
int nhum;

int prime[MAXPRIME];
int pindex[MAXPRIME];
int nprime;

void
main(void)
{
FILE *fin, *fout;
int i, minp;
long min;
int n;

fin = fopen("humble.in", "r");
fout = fopen("humble.out", "w");
assert(fin != NULL && fout != NULL);

fscanf(fin, "%d %d", &nprime, &n);
for(i=0; i<nprime; i++)
fscanf(fin, "%d", &prime[i]);

hum[nhum++] = 1;
for(i=0; i<nprime; i++)
pindex[i] = 0;

while(nhum < n+1) {
min = 0x7FFFFFFF;
minp = -1;
for(i=0; i<nprime; i++) {
while((double)prime[i] * hum[pindex[i]] <= hum[nhum-1])
pindex[i]++;

/* double to avoid overflow problems */
if((double)prime[i] * hum[pindex[i]] < min) {
min = prime[i] * hum[pindex[i]];
minp = i;
}
}

hum[nhum++] = min;
pindex[minp]++;
}

fprintf(fout, "%d\n", hum[n]);
exit(0);
}
``````