Humble Numbers


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22427    Accepted Submission(s): 9800



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.


 


HDU 1058 Humble Numbers(打表+暴力)_#define




题解：


找出能被2，3，5，7整除的数，即素因子。既是素数，又是因子。然后找出是第几个。。。。




AC代码：


暴力+打表




#include <cstdio>
#define min(a,b) ((a) < (b) ? (a):(b))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
int a[5850];
int main(void)
{
  int n = 1;
  int p2,p3,p5,p7;
  p2 = p3 = p5 = p7 = 1;
  a[1] = 1;
  while(a[n] < 2000000000)
  {
    a[++n] = min4(2 * a[p2],3 * a[p3], 5 * a[p5], 7 * a[p7]);
    if(a[n] == 2 * a[p2])
      p2++;
    if(a[n] == 3 * a[p3])
      p3++;  
    if(a[n] == 5 * a[p5])
      p5++;
    if(a[n] == 7 * a[p7])
      p7++;
  }
  while(~scanf("%d",&n))
  {
    if(n==0)break;
    printf("The %d",n);
    int ten = n/10%10;
    if(n%10 == 1 && ten != 1)
      printf("st");
    else if(n%10 == 2 && ten != 1)
      printf("nd");
    else if(n%10 == 3 && ten != 1)
      printf("rd");
    else
      printf("th");
    printf(" humble number is %d.\n",a[n]);
  }
  return 0;
}