HDU 1058 Humble Numbers【巧用优先队列】

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最新推荐文章于 2020-10-31 21:33:16 发布

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本文探讨如何通过优先队列、集合数据结构和预处理技术，在限制时间内找到指定位置的谦逊数，并且详细解释了输出格式的特殊要求。

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21717 Accepted Submission(s): 9480

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.

见到这个题，直接想到了之前做的一道题的处理方法，奈何猜到了开头，没猜到中间，更没猜到结局..........

三个点：

第一，用优先队列和 set 来生成符合要求的数列

第二，需要打表来保存数据，否则会超时

第三，注意输出的格式的控制（第一次见到，涨姿势了....）

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
ll num[6005];
void db()
{
	int x[4]={2,3,5,7},cnt=1;
	priority_queue<ll,vector<ll>,greater<ll> > q;
	set<ll> set;
	q.push(1);set.insert(1);
	while(cnt<6000)
	{
		ll tp=q.top(),next;q.pop();
		num[cnt]=tp;
		for(int i=0;i<4;++i)
		{
			next=x[i]*tp;
			if(!set.count(next))
			{
				set.insert(next);
				q.push(next);
			}
		}
		++cnt;
	}
}
int main()
{
	db();
	int n;
	while(scanf("%d",&n),n)
	{
		printf("The %d",n);
		if(n%10==1&&n%100!=11)
		{
			printf("st ");
		}
		else if(n%10==2&&n%100!=12)
		{
			printf("nd ");
		}
		else if(n%10==3&&n%100!=13)
		{
			printf("rd ");
		}
		else 
		{
			printf("th ");
		}
		printf("humble number is %lld.\n",num[n]);
	}
	return 0;
}