[3_1_humble] maintain many pointers & scan

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本文介绍了如何找到给定质数集合中的第N个‘谦逊数’，通过实现算法并提供输入输出格式示例，帮助理解如何计算并输出这些特殊的数。

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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)

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);
}