Ugly Numbers--POJ 1338

1、解题思路：数论、STL。

2、注意事项：i++使用，跟踪i的值；STL的简单应用。

3、实现方法：（复杂版）

  
    

   
     
   
      1 
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
      2 
   
     using
   
      
   
     namespace
   
      std;

   
      3 
   
     

   
      4 
   
     int
   
      UglyNumbers[
   
     1500
   
     ];

   
      5 
   
     

   
      6 
   
     int
   
      MinNumber(
   
     int
   
      a,
   
     int
   
      b,
   
     int
   
      c)

   
      7 
   
     {

   
      8 
   
      
   
     int
   
      min
   
     =
   
     a; 

   
      9 
   
      
   
     if
   
     (min
   
     >
   
     b) min
   
     =
   
     b;

   
     10 
   
      
   
     if
   
     (min
   
     >
   
     c) min
   
     =
   
     c; 

   
     11 
   
      
   
     return
   
      min;

   
     12 
   
     }

   
     13 
   
     

   
     14 
   
     void
   
      Init()

   
     15 
   
     {

   
     16 
   
      
   
     int
   
      two
   
     =
   
     0
   
     ,three
   
     =
   
     0
   
     ,five
   
     =
   
     0
   
     ,temp;

   
     17 
   
      UglyNumbers[
   
     0
   
     ]
   
     =
   
     1
   
     ;

   
     18 
   
      
   
     for
   
     (
   
     int
   
      i
   
     =
   
     1
   
     ;i
   
     <
   
     1500
   
     ;i
   
     ++
   
     )

   
     19 
   
      {

   
     20 
   
      temp
   
     =
   
     MinNumber(UglyNumbers[two]
   
     *
   
     2
   
     ,UglyNumbers[three]
   
     *
   
     3
   
     ,UglyNumbers[five]
   
     *
   
     5
   
     );

   
     21 
   
      
   
     if
   
     (temp
   
     ==
   
     UglyNumbers[two]
   
     *
   
     2
   
     )

   
     22 
   
      two
   
     ++
   
     ;

   
     23 
   
      
   
     if
   
     (temp
   
     ==
   
     UglyNumbers[three]
   
     *
   
     3
   
     )

   
     24 
   
      three
   
     ++
   
     ;

   
     25 
   
      
   
     if
   
     (temp
   
     ==
   
     UglyNumbers[five]
   
     *
   
     5
   
     ) 

   
     26 
   
      five
   
     ++
   
     ;

   
     27 
   
      UglyNumbers[i]
   
     =
   
     temp;

   
     28 
   
      }

   
     29 
   
     }

   
     30 
   
     

   
     31 
   
     int
   
      seach(
   
     int
   
      x)

   
     32 
   
     {

   
     33 
   
      
   
     return
   
      UglyNumbers[x
   
     -
   
     1
   
     ];

   
     34 
   
     }

   
     35 
   
     

   
     36 
   
     int
   
      main() 

   
     37 
   
     {

   
     38 
   
      
   
     int
   
      n;

   
     39 
   
      Init();

   
     40 
   
      
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n)

   
     41 
   
      {

   
     42 
   
      cout
   
     <<
   
     seach(n)
   
     <<
   
     endl;

   
     43 
   
      } 

   
     44 
   
      
   
     return
   
      
   
     0
   
     ;

   
     45 
   
     }

   
     46 
  
    

4、实现方法：（STL精简版）

  
    

   
     
   
      1 
   
     #include
   
     <
   
     algorithm
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
      3 
   
     #include
   
     <
   
     queue
   
     >
   
     

   
      4 
   
      
   
     using
   
      
   
     namespace
   
      std;

   
      5 
   
     typedef pair
   
     <
   
     unsigned 
   
     long
   
     ,
   
     int
   
     >
   
      node_type;

   
      6 
   
     

   
      7 
   
     int
   
      main()

   
      8 
   
     { 

   
      9 
   
      unsigned 
   
     long
   
      result[
   
     1500
   
     ]; 

   
     10 
   
      priority_queue
   
     <
   
      node_type, vector
   
     <
   
     node_type
   
     >
   
     , greater
   
     <
   
     node_type
   
     >
   
      
   
     >
   
      Q; 

   
     11 
   
      Q.push( make_pair(
   
     1
   
     ,
   
     2
   
     ) ); 

   
     12 
   
      
   
     for
   
     (
   
     int
   
      i
   
     =
   
     0
   
     ; i
   
     <
   
     1500
   
     ; i
   
     ++
   
     ) 

   
     13 
   
      { 

   
     14 
   
      node_type node
   
     =
   
     Q.top(); 

   
     15 
   
      Q.pop(); 

   
     16 
   
      
   
     switch
   
     (node.second) 

   
     17 
   
      { 

   
     18 
   
      
   
     case
   
      
   
     2
   
     :Q.push( make_pair(node.first
   
     *
   
     2
   
     ,
   
     2
   
     )); 

   
     19 
   
      
   
     case
   
      
   
     3
   
     :Q.push( make_pair(node.first
   
     *
   
     3
   
     ,
   
     3
   
     )); 

   
     20 
   
      
   
     case
   
      
   
     5
   
     :Q.push( make_pair(node.first
   
     *
   
     5
   
     ,
   
     5
   
     )); 

   
     21 
   
      } 

   
     22 
   
      result[i]
   
     =
   
     node.first; 

   
     23 
   
      } 

   
     24 
   
      
   
     int
   
      n; 
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n) 

   
     25 
   
      { 

   
     26 
   
      cout
   
     <<
   
     result[n
   
     -
   
     1
   
     ]
   
     <<
   
     endl; 

   
     27 
   
      } 

   
     28 
   
      
   
     return
   
      
   
     1
   
     ;}