Hdoj 1058.Humble Numbers 题解

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

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

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.

Source

University of Ulm Local Contest 1996

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思路

丑数的因子只有2,3,5,7，所以可以用从1开始乘这四个因子得到新的数字之后加到数组里面，更新1为2，如此迭代计算下去到算出5842个值，但是这样会存在重复元素问题还要考虑去重问题，数组的元素也不是有序的。

改善这个思路可以有：

设数组a存放所有的丑数，设立4个游标pos分别对应4个因子2,3,5,7，一个数组暂时存储所得的结果。每次都用游标值*4个因子，然后找到最小的值放入数组a，直到数量为5842为止。

注意一个坑：111的英文是one hundred and eleventh，是th结尾的！！！！！！，112，113同理，实际上不止是这三个数，只要符合一定的特征。

代码

#include<bits/stdc++.h>
using namespace std;
int num[6000];


int findMin(int a[],bool f[],int len)
{
	int minvalue = a[0];
	for(int i=0;i<len;i++)
		minvalue = min(minvalue, a[i]);
	
	for(int i=0;i<len;i++)
		minvalue == a[i] ? f[i] = true : f[i] = false;
	return minvalue;
}

void Init()
{
	num[1] = 1;
	bool vis[6000];
	int tmp[4];
	int pos[4] = {1,1,1,1};
	int fac[4] = {2,3,5,7};
	
	for(int i=2;i<=5842;i++)
	{
		for(int j=0;j<4;j++)
			tmp[j] = num[pos[j]] * fac[j];
		
		int minvalue = findMin(tmp,vis,4);
		num[i] = minvalue;
		for(int j=0;j<4;j++)
			if(vis[j])
				pos[j]++;
	}
}
int main()
{
	int n;
	Init();
	while(cin>>n && n!=0)
	{
		string suffix = "th";
		if(n%10==1 && n%100!=11)  
            suffix = "st";  
        else if(n%10==2 && n%100!=12)  
            suffix = "nd";  
        else if(n%10==3 && n%100!=13)  
            suffix = "rd";  
  
		cout<<"The "<<n<<suffix<<" humble number is "<<num[n]<<"."<<endl; 
    }
	return 0;
}