NOIP: Humble Number

最新推荐文章于 2020-04-13 09:19:16 发布

purfan2

最新推荐文章于 2020-04-13 09:19:16 发布

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本文介绍了一种特殊的数列——谦逊数序列，这些数仅由2、3、5或7的质因数组成。通过一个高效的算法实现，文章详细解释了如何找出序列中的第n个元素，并提供了一个C++代码示例，用于计算特定位置上的谦逊数。

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描述

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.
输入
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.
输出
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.
样例输入
1
2
3
4
11
12
13
21
22
23
100
1000
5842
0
样例输出
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.

代码

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

int dp[6000];

int main()
{
	dp[1] = 1;
	// 定义计数器 
 	int c2 = 1, c3 = 1, c5 = 1, c7 = 1;
	for(int i = 2; i <= 5842; i++)
	{
		int v2 = dp[c2] * 2;
		int v3 = dp[c3] * 3;
		int v5 = dp[c5] * 5;
		int v7 = dp[c7] * 7;
		int ans = min(min(min(v2,v3),v5),v7);
		dp[i] = ans;
		if(ans == v2)
		{
			c2++;
		}
		if(ans == v3)
		{
			c3++;
		}
		if(ans == v5)
		{
			c5++;
		}
		if(ans == v7)
		{
			c7++;
		}		
	}
	
	
	int n;
	while(scanf("%d", &n)!=EOF && n!=0) 
	{	
		cout<<"The "<<n;		
		if(n % 100 == 11 || n % 100 == 12 || n % 100== 13) 
		{
			cout<<"th";
		}
		else if(n % 10 == 1) 
		{
			cout<<"st";
		}
		else if(n % 10 == 2) 
		{
			cout<<"nd";
		}		 
		else if(n % 10 == 3) 
		{
			cout<<"rd";
		}	
		else
		{
			cout<<"th";
		}
		cout<<" humble number is "<<dp[n]<<"."<<endl;
	}
	return 0;	
}