HDOJ1058 Humble Numbers丑数的问题

最新推荐文章于 2019-08-31 21:39:40 发布

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本文介绍了一种特殊的数列——谦逊数（Humble Numbers），这些数仅由2、3、5、7作为质因数构成。通过动态规划的方法，文章详细解释了如何高效地找出数列中的指定元素，并附带提供了完整的C++实现代码。

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24222 Accepted Submission(s): 10610

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

状态转移方程：F(n)=min(F(i)*2,F(j)*3,F(k)*5,F(m)*7)

(n>i,j,k,m)

特别的：
i,j,k,m 只有在本项被选中后才移动

ijkm的数值表示在第该数值的答案是由ijkm个对应因素乘过的数字。

Sample Output

The 1st humble #include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int ugly[5842];

int Min(int a,int b,int c,int d)
{
	int temp=(a<b?a:b);
	temp= (temp<c?temp:c);
	
	return (temp<d?temp:d);
}


int main()
{

	int N;
	
	ugly[0]=1;

	int index2=0;
	int index3=0;
	int index5=0;
	int index7=0;
	
	int index=1;
	
	while(index<5842)
	{
			int val=Min(ugly[index2]*2,ugly[index3]*3,ugly[index5]*5,ugly[index7]*7);
			
			if(val==ugly[index2]*2)
				index2++;
			if(val==ugly[index3]*3)
				index3++;
			if(val==ugly[index5]*5)
				index5++;
			if(val==ugly[index7]*7)
				index7++;
				
			ugly[index++]=val;
	}
		
	while(scanf("%d",&N)&&N!=0)
	{
		
		if(N % 10 == 1 && N % 100 != 11)
			printf("The %dst humble number is %d.\n",N,ugly[N-1]);
		else if(N % 10 == 2 && N % 100 != 12)
			printf("The %dnd humble number is %d.\n",N,ugly[N-1]);
		else if(N % 10 == 3 && N % 100 != 13)
			printf("The %drd humble number is %d.\n",N,ugly[N-1]);
		else 
			printf("The %dth humble number is %d.\n",N,ugly[N-1]);
	}
	
	
	return 0;
}

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.