USACO3.1.3Humble 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
题解：自己写了个好暴力的程序，第四个点就超时了。。。木有想到怎么优化，只好看题解，照着题解翻译的，好失败%>_<%。直接贴官方题解好了。。

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.

/*
ID:spcjv51
PROG:humble
LANG:C
*/
#include<stdio.h>
#include<string.h>
long a[105];
int f[105];
long num[100005];
int main(void)
{
    freopen("humble.in","r",stdin);
    freopen("humble.out","w",stdout);
    long i,k,n,ans,min;
    scanf("%ld%ld",&n,&k);
    memset(f,0,sizeof(f));
    num[0]=1;
    for(i=1;i<=n;i++)
        scanf("%ld",&a[i]);
    ans=0;
    while(ans<k)
    {
        min=0x7FFFFFFF;
        for(i=1;i<=n;i++)
        {
            while(a[i]*num[f[i]]<=num[ans])
            f[i]++;
            if(a[i]*num[f[i]]<min) min=a[i]*num[f[i]];
        }
        ans++;
        num[ans]=min;
    }
    printf("%ld\n",num[k]);
    return 0;
}