POJ 1850 Code

解题思路：组合数学题，注意题目要求升序排列

类似于POJ3252，c[i][j]存储(choose i from j 的)组合值

  
    

   
     
   
      1 
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
      2 
   
      
   
     using
   
      
   
     namespace
   
      std;

   
      3 
   
     int
   
      main()

   
      4 
   
     {

   
      5 
   
      
   
     int
   
      ans,l,i,j,p,q,c[
   
     27
   
     ][
   
     27
   
     ]
   
     =
   
     {
   
     0
   
     },nc[
   
     11
   
     ],IsOk
   
     =
   
     1
   
     ;

   
      6 
   
      
   
     char
   
      ch[
   
     11
   
     ];

   
      7 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     27
   
     ;i
   
     ++
   
     )

   
      8 
   
      
   
     for
   
     (c[i][i]
   
     =
   
     1
   
     ,j
   
     =
   
     i
   
     +
   
     1
   
     ;j
   
     <
   
     27
   
     ;j
   
     ++
   
     )

   
      9 
   
      c[i][j]
   
     =
   
     c[i][j
   
     -
   
     1
   
     ]
   
     *
   
     j
   
     /
   
     (j
   
     -
   
     i);

   
     10 
   
      scanf(
   
     "
   
     %s
   
     "
   
     , ch);l
   
     =
   
     strlen(ch);

   
     11 
   
      
   
     for
   
     (ans
   
     =
   
     i
   
     =
   
     0
   
     ;i
   
     <
   
     l
   
     &&
   
     IsOk;i
   
     ++
   
     ){nc[i
   
     +
   
     1
   
     ]
   
     =
   
     ch[i]
   
     -
   
     '
   
     a
   
     '
   
     +
   
     1
   
     ;
   
     if
   
     (ch[i]
   
     >
   
     '
   
     z
   
     '
   
     ||
   
     ch[i]
   
     <
   
     '
   
     a
   
     '
   
     ||
   
     nc[i
   
     +
   
     1
   
     ]
   
     <=
   
     nc[i])IsOk
   
     =
   
     0
   
     ;}

   
     12 
   
      
   
     if
   
     (IsOk)

   
     13 
   
      {

   
     14 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     l;i
   
     ++
   
     )

   
     15 
   
      ans
   
     +=
   
     c[i][
   
     26
   
     ];

   
     16 
   
      
   
     for
   
     (p
   
     =
   
     0
   
     ,i
   
     =
   
     1
   
     ,q
   
     =
   
     nc[i];i
   
     <=
   
     l;i
   
     ++
   
     ,p
   
     =
   
     nc[i
   
     -
   
     1
   
     ],q
   
     =
   
     nc[i])

   
     17 
   
      
   
     for
   
     (j
   
     =
   
     27
   
     -
   
     q;j
   
     <
   
     26
   
     -
   
     p;j
   
     ++
   
     )ans
   
     +=
   
     c[l
   
     -
   
     i][j];

   
     18 
   
      ans
   
     ++
   
     ;

   
     19 
   
      }

   
     20 
   
      printf(
   
     "
   
     %d\n
   
     "
   
     ,ans);

   
     21 
   
      
   
     return
   
      
   
     0
   
     ;

   
     22 
   
     }