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最新推荐文章于 2024-07-27 08:45:00 发布

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本文介绍了一种特殊的数列——谦逊数（Humble Number），并提供了一种算法来找出数列中的任意一项。谦逊数是指仅包含2、3、5、7作为质因数的整数。文章通过示例解释了如何生成这些数，并给出了一段C语言代码实现。

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.

分析：若一个数是Humble数，则它的2、3、5、7倍仍然是Humble数。
设a[i]为第i个Humble数，则a[n] = min(2*a[b2], 3*a[b3], 5*a[b5], 7*a[b7]),
b2、b3、b5、b7在不断更新
*/

#include<stdio.h>
int Min(int a, int b, int c, int d)//找到a,b,c,d中小的那个数 
{    int Min1 = a< b ? a : b;   
     int Min2 = c< d ? c : d;   
  return Min1< Min2 ? Min1 : Min2;
  }
int main()
{    int n, a[5850];
     int b2, b3, b5, b7; 
     int m ; 
     int temp;
        m=1; 
        a[1] = 1;    
        b2 = b3 = b5 = b7 = 1;   
	 while(m<= 5842) //打表，从最小的开始往上增加 
	{       
	   temp = Min(2*a[b2], 3*a[b3], 5*a[b5], 7*a[b7]);    
	      a[++m] = temp;     
	    if(temp == 2*a[b2]) b2++;      
		if(temp == 3*a[b3]) b3++;     
		if(temp == 5*a[b5]) b5++;  
		if(temp == 7*a[b7]) b7++; 
   }  
   	while(~scanf("%d",&n) && n)  
	{     
	 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",a[n]);   
			}   
						 
		return 0;
		 }