Humble numbers HDU 1058 ACM

Humble numbers

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 6085 accepted submission (s): 2639

Problem descriptiona 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

Inputthe 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.

Outputfor 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 input1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

Sample outputthe 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.

Sourceuniversity of Ulm Local contest 1996

Recommendjgshining

 /*  
The idea for this question is very simple: we use, loops.
Find the I th f [I], and the I th f [I] must be equal
One of the preceding I-1 numbers corresponds to {2, 3, 5, 7}
The product of one of them, so the answer comes out ~~


 */  
# Include  <  Stdio. h  >  
# Include  <  Iostream  >  
  Using     Namespace  STD;
  Long     Long F [  5843  ];
  Int  Main ()
{
  Int  Prime [  4  ]  =  {  2  ,  3  ,  5  ,  7 };
  Int  I, J, K;
  Int  N;
F [  1  ]  =  1  ;
  For  (I  =  2  ; I  <=  5842  ; I ++  )
{
F [I]  =  2000000001  ;
  For  (J  =  0  ; J  <  4  ; J  ++  )
{
  For  (K =  I  -  1  ; K  > =  1  ; K  --  )
{
  If  (F [k]  *  Prime [J]  <=  F [I  -  1 ])  Break  ;
  If  (F [k]  *  Prime [J]  <  F [I])
F [I]  =  F [k]  *  Prime [J];
}
}
}
  While  (CIN  > N, N)
{
Cout  <  "  The  "  <  N;
  If  (N  %  10  =  1  &&  N  %  100 ! =  11  ) Cout  <  "  St  "  ;
  Else     If  (N  %  10  =  2  &&  N %  100  ! =  12  ) Cout  <  "  Nd  "  ;
  Else     If  (N  %  10  =  3 &&  N  %  100  ! =  13  ) Cout  <  "  Rd  "  ;
  Else  Cout  <  "  Th  " ;
Cout  <  "  Humble number is  "  <  F [N]  <  "  .  "  <  Endl;

}
  Return     0  ;
}