poj2623

快排，注意保留一位小数。

View Code

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstdio
    
      >
    
      
#include 
    
      <
    
      cstdlib
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      
#include 
    
      <
    
      algorithm
    
      >
    
      
#include 
    
      <
    
      cmath
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      #define
    
       maxn 250005
    
      


    
      int
    
       n;

    
      long
    
       
    
      long
    
       f[maxn], sum;


    
      int
    
       main()
{
 
    
      //
    
      freopen("t.txt", "r", stdin);
    
      

    
       scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n);
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 scanf(
    
      "
    
      %lld
    
      "
    
      , 
    
      &
    
      f[i]);
 sort(f, f 
    
      +
    
       n);
 
    
      if
    
       (n 
    
      &
    
       
    
      1
    
      )
 printf(
    
      "
    
      %lld.0\n
    
      "
    
      , f[n 
    
      /
    
       
    
      2
    
      ]);
 
    
      else
    
      
 {
 sum 
    
      =
    
       
    
      0
    
      ;
 sum 
    
      +=
    
       f[n 
    
      /
    
       
    
      2
    
      ];
 sum 
    
      +=
    
       f[n 
    
      /
    
       
    
      2
    
       
    
      -
    
       
    
      1
    
      ];
 
    
      if
    
       (sum 
    
      &
    
       
    
      1
    
      )
 printf(
    
      "
    
      %lld.5\n
    
      "
    
      , sum 
    
      /
    
       
    
      2
    
      );
 
    
      else
    
      
 printf(
    
      "
    
      %lld.0\n
    
      "
    
      , sum 
    
      /
    
       
    
      2
    
      );
 }
 
    
      return
    
       
    
      0
    
      ;
}