HDOJ-1058 Humble Numbers（动态规划）

版权声明：个人学习笔记记录 https://blog.csdn.net/Ratina/article/details/84936900

链接：HDOJ-1058

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.

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题目大意：

因子只有2，3，5，7的数称为丑数（1也是丑数），找到第N个丑数（N<=5842）

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分析：

思路不难想，对已有的所有数*2，*3，*5，*7，从中选出能放在下一位的数。
但是知道了这个却一直想不出好的dp方法。。。 后来看到别人的状态转移方程豁然开朗。。。
如下：

dp[i] = min( dp[two] * 2 , dp[three] * 3 , dp[five] * 5 , dp[seven] * 7 )

two，three，five，seven是*2，*3，*5，*7 进行到第几个数的位置标记，这样保证了已有的所有数都进行 *2，*3，*5，*7而且只进行一次，同时进行min比对。
得出的 dp[i] 会等于其中某一个结果（或多个结果：*2，*3，*5，*7可能有相同值），就让对应位置标记+1，这样同时能保证进行min对比的都是相对小的值。

此外，就是这题输出的坑，结尾为1，2，3并且结尾不是11，12，13的N后面要加上"st"，“nd”，“rd”，其他的加"th"。

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以下代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int dp[6000];
	int i,two=1,three=1,five=1,seven=1;
	dp[1]=1;
	for(i=2;i<=5842;i++)
	{
		dp[i]=min(min(dp[two]*2,dp[three]*3),min(dp[five]*5,dp[seven]*7));
		if(dp[i]==dp[two]*2)
			two++;
		if(dp[i]==dp[three]*3)
			three++;
		if(dp[i]==dp[five]*5)
			five++;
		if(dp[i]==dp[seven]*7)
			seven++;
	}
	int N;
	while(scanf("%d",&N)&&N!=0)
	{
		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 %d.\n",dp[N]);
	}
	return 0;
}