HDU 1058 Humble Numbers（丑数，优先队列，STL）

本文介绍了一种通过优先队列和集合数据结构高效求解Humble Numbers的方法，并给出了完整的AC代码实现。Humble Numbers是指仅包含2、3、5、7作为质因数的整数。

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http://acm.hdu.edu.cn/showproblem.php?pid=1058

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24358 Accepted Submission(s): 10675

Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.

Write a program to find and print the nth element in this sequence

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.

Sample Input

Sample Output

  
   The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

题意：

丑数的定义（这里不同于数论的丑数定义），本题目类似于 UVA136 （点击链接到）。

该题目丑数是只含有素因数为2,3,5,7的数。键入 n 值，输出第 n 个丑数。

思路：

最小的丑数是 1，而对于任意丑数 x ，有2x, 3x, 5x, 7x也都是丑数。此时使用一个优先队列保存已经生成丑数中的最小丑数，再次生成 4 个丑数....

注意的是需要判断丑数是否已经生成。

注：数字后的英语字母！！！

AC CODE：

#include<stdio.h>
#include<cstring>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
typedef long long LL;
using namespace std;
const int MYDD = 1103+5842;

LL ans[MYDD];
int base[4] = {2, 3, 5, 7};
void Init() {
	priority_queue< LL, vector<LL>, greater<LL> > pq;
	set<LL> s;
	pq.push(1);
	s.insert(1);
	for(int j = 1; j <= 5842; j++) {
		LL x = pq.top();
		ans[j] = x;
		pq.pop();
		for(int j = 0; j < 4; j++) {
			LL x2 = x*base[j];
			if(!s.count(x2)) {
				s.insert(x2);
				pq.push(x2);
			}
		}
	}
}

int HardBoy {
	Init();
	int n;
	while(scanf("%d", &n) && n) {
//		printf("The %dth humble number is %d.\n", n, ans[n]);
		if(n%10==1&&n%100!=11)
			printf("The %dst humble number is %d.\n", n, ans[n]);
		else if(n%10==2&&n%100!=12)
			printf("The %dnd humble number is %d.\n", n, ans[n]);
		else if(n%10==3&&n%100!=13)
			printf("The %drd humble number is %d.\n", n, ans[n]);
		else
			printf("The %dth humble number is %d.\n", n, ans[n]);
	}
	ForMyLove
}