USACO: 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

                          

思路：

一个新的humble number可以从已得到的序列得到：H(n+1) = min{ H(i) * prime(j) > H(n), 0 <= i <= n, 0 <= j < k }

显然，对于prime[x]，humb中有一个对应的起始位置nxpos[x]（上一次找到这里发现H(nxpos[x]) * prime[x] > H(n + 1)），那么下次可以接着往下找，而不是从头开始找

/* ID: geterns1 PROG: humble LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; const int MAXP = 100; const int MAX = 100000; int cnt = 0; int prime[MAXP], k; //keep the primes int nxpos[MAXP]; //nxpos[i] keep the next position should prime[i] begin in humb long humb[MAX + 1]; int n; //keep the humble number order 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", prime + i); humb[0] = 1; //trade 1 as the first humble number for (int i = 0; i < k; i ++) nxpos[i] = 0; long min; while (cnt < n){ min = 0x7fffffff; for (int i = 0; i < k; i ++){ while ((double)prime[i] * humb[nxpos[i]] <= humb[cnt]) //overflow may occure in long integer nxpos[i] ++; if ((double)prime[i] * humb[nxpos[i]] < min) min = prime[i] * humb[nxpos[i]]; } humb[++ cnt] = min; } printf("%d/n", humb[n]); fclose(stdin); fclose(stdout); return 0; }