hdu 1058 Humble Numbers

B - Humble Numbers

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1058

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.
      
      
      
      
      
      
      
      


      
      
      
      
      
      
      
      


      
      
      
      
      
      
      
      

       
       
       
       
这道题开始做的时候似懂非懂，如何摸清楚dp[i]与前面的数的关系是关键。
      
      
      
      
      
      
      
      

       
       
       
       我们非常容易知道dp[i]=dp[a]*2||dp[b]*3||dp[c]*5||dp[d]*7。但是我们难以确定到底是a,b,c,d是哪一个数。
      
      
      
      
      
      
      
      

       
       
       
       容易发现对于dp[j],它乘以2或3或5或7产生的每一个数都只能存入dp数组一次。
      
      
      
      
      
      
      
      

       
       
       
       即可的到状态转移方程：
      
      
      
      
      
      
      
      

       
       
       
       dp[i]=min(dp[a]*2,min(dp[b]*3,min(dp[c]*5,dp[d]*7)));
      
      
      
      
      
      
      
      


      
      
      
      
      
      
      
      


      
      
      
      
      
      
      
      

       
       
       
       
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int dp[6000];
int main()
{
	freopen("in.txt","r",stdin);
	dp[1]=1;
	int a,b,c,d;
	a=b=c=d=1;
	for(int i=2;i<=5842;i++)
	{
		dp[i]=min(dp[a]*2,min(dp[b]*3,min(dp[c]*5,dp[d]*7)));
		if(dp[i]==dp[a]*2) a++;
		if(dp[i]==dp[b]*3) b++;
		if(dp[i]==dp[c]*5) c++;
		if(dp[i]==dp[d]*7) d++;
	}
	int n;
	while(cin>>n)
	{
		if(n==0) break;
		if(n%100==11||n%100==12||n%100==13)
			printf("The %dth humble number is %d.\n",n,dp[n]);
		else if(n%10==1)
			printf("The %dst humble number is %d.\n",n,dp[n]);
		else if(n%10==2)
			printf("The %dnd humble number is %d.\n",n,dp[n]);
		else if(n%10==3)
			printf("The %drd humble number is %d.\n",n,dp[n]);
		else
			printf("The %dth humble number is %d.\n",n,dp[n]);
	}
	return 0;
}