USACO3.1.3--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
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.

View Code

 1 /*

 2 ID:spcjv51

 3 PROG:humble

 4 LANG:C

 5 */

 6 #include<stdio.h>

 7 #include<string.h>

 8 long a[105];

 9 int f[105];

10 long num[100005];

11 int main(void)

12 {

13     freopen("humble.in","r",stdin);

14     freopen("humble.out","w",stdout);

15     long i,k,n,ans,min;

16     scanf("%ld%ld",&n,&k);

17     memset(f,0,sizeof(f));

18     num[0]=1;

19     for(i=1;i<=n;i++)

20         scanf("%ld",&a[i]);

21     ans=0;

22     while(ans<k)

23     {

24         min=0x7FFFFFFF;

25         for(i=1;i<=n;i++)

26         {

27             while(a[i]*num[f[i]]<=num[ans])

28             f[i]++;

29             if(a[i]*num[f[i]]<min) min=a[i]*num[f[i]];

30         }

31         ans++;

32         num[ans]=min;

33     }

34     printf("%ld\n",num[k]);

35     return 0;

36 }