COJ 1016 Simple calculations 难度在于推导求解a1的公式，编程很简单

Simple calculations

Description

There is a sequence of n+2 elements a ₀, a ₁, ..., a _n+1 (n <= 3000, -1000 <= a _i <=1000). It is known that a_i = (a_i-1 + a_i+1)/2 - c_i for each i=1, 2, ..., n.
You are given a ₀, a _n+1, c ₁, ... , c _n. Write a program which calculates a ₁.

Input

The first line of an input contains an integer n. The next two lines consist of numbers a ₀ and a _n+1 each having two digits after decimal point, and the next n lines contain numbers c _i (also with two digits after decimal point), one number per line.

Output

The output file should contain a ₁ in the same format as a ₀ and a _n+1.

Sample Input

1 50.50 25.50 10.15

Sample Output

27.85

题意：由公式a_i = (a_i-1 + a_i+1)/2 - c_{i ；}

_给出 a₀ a_{n+1 和}c_1，c_2，c_3......c_i,求出a₁

方法：

 2a_n = a_n-1 + a_n+1 - 2c_{n ；等价于}_{a_{n+1 -}}_{a_n = a_n - a_n-1 + 2c_n}

2a_n-1 = a_n-2 + a_n - 2c_{n-1 ；}_等价于_{a_{n -}}_{a_n-1 = a_n-1 - a_n-2 + 2c_n}

 . . . .

 . . . .

 . . . .



2a₃ =  a₂
  +  a₄
  - 2c_{3 ；_等价于_{a_{4 -}}_{a₃ = a₃ - a₂ + 2c₃}}

2a₂ =  a₁
  +  a₃
  - 2c_{2 ；}_{_等价于_{a_{3 -}}_{a₂ = a₂ - a₁ + 2c₂}}

 2a₁ = a₀ + a₂ - 2c1_；_{_{_等价于_{a_{2 -}}_{a₁ = a₁ - a₀ + 2c₁}}}

_{相加得出：}

a_n+1 - a₁ = a_n-a₀+ 2(c1+c2+c3+....+cn)

因为 
          
            
                
                 a_n
  
               
 
  
         
 还不知道，所以要转化：

a_n+1 - a_n = a₁-a₀+ 2(c1+c2+c3+..............+cn)

 a_n - a_n-1 = a₁-a₀+ 2(c1+c2+c3+..........+c_n-1)

 a_n-1 - a_n-2 = a₁-a₀+ 2(c1+c2+c3+....+c_n-2)

 . . . . . . .

 . . . . . . .

 . . .. . . . . 





     
     
     
      
       a₃ - a₂ = a₂-a₁+ 2(c1 + 
  
     


     
     
     
      
       

         
         
         
          
           c2)

 a₂ - a₁ = a₁-a₀+ 2c1

相加化简得：

 a_n+1 - a₁= a₁-a₀+2( nc₁+ (n-1)*c₂+ (n-2)c₃+.....+2c_n-1+c_n )

_化简得：

a₁=(a_n+1+na₀-2( nc₁+ (n-1)*c₂+ (n-2)c₃+.....+2c_n-1+c_n ))/(n+1)


    
    
    
    
     
     
     
     
      1 
     
     
     
     #include 
     
     
     
     <
     
     
     
     cstdlib
     
     
     
     >
     
     
     
     

     
     
     
      2 
     
     
     
     #include 
     
     
     
     <
     
     
     
     iostream
     
     
     
     >
     
     
     
     

     
     
     
      3 
     
     
     
     

     
     
     
      4 
     
     
     
     
     
     
     
     using
     
     
     
      
     
     
     
     namespace
     
     
     
      std;

     
     
     
      5 
     
     
     
     
     
     
     
     float
     
     
     
      c[
     
     
     
     3001
     
     
     
     ],a[
     
     
     
     3002
     
     
     
     ];

     
     
     
      6 
     
     
     
     
     
     
     
     int
     
     
     
      main(
     
     
     
     int
     
     
     
      argc, 
     
     
     
     char
     
     
     
      
     
     
     
     *
     
     
     
     argv[])

     
     
     
      7 
     
     
     
     {

     
     
     
      8 
     
     
     
         
     
     
     
     int
     
     
     
      n,i,j; 
     
     
     
     float
     
     
     
      cc
     
     
     
     =
     
     
     
     0
     
     
     
     ;

     
     
     
      9 
     
     
     
         cin
     
     
     
     >>
     
     
     
     n;

     
     
     
     10 
     
     
     
         cin
     
     
     
     >>
     
     
     
     a[
     
     
     
     0
     
     
     
     ]
     
     
     
     >>
     
     
     
     a[n
     
     
     
     +
     
     
     
     1
     
     
     
     ];

     
     
     
     11 
     
     
     
         
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     1
     
     
     
     ;i
     
     
     
     <=
     
     
     
     n;i
     
     
     
     ++
     
     
     
     )

     
     
     
     12 
     
     
     
         {

     
     
     
     13 
     
     
     
             cin
     
     
     
     >>
     
     
     
     c[i];

     
     
     
     14 
     
     
     
             cc
     
     
     
     +=
     
     
     
     (n
     
     
     
     -
     
     
     
     i
     
     
     
     +
     
     
     
     1
     
     
     
     )
     
     
     
     *
     
     
     
     c[i];

     
     
     
     15 
     
     
     
         }

     
     
     
     16 
     
     
     
         a[
     
     
     
     1
     
     
     
     ]
     
     
     
     =
     
     
     
     (a[n
     
     
     
     +
     
     
     
     1
     
     
     
     ]
     
     
     
     +
     
     
     
     n
     
     
     
     *
     
     
     
     a[
     
     
     
     0
     
     
     
     ]
     
     
     
     -
     
     
     
     2
     
     
     
     *
     
     
     
     cc)
     
     
     
     /
     
     
     
     (n
     
     
     
     +
     
     
     
     1
     
     
     
     );

     
     
     
     17 
     
     
     
         printf(
     
     
     
     "
     
     
     
     %.2f
     
     
     
     "
     
     
     
     ,a[
     
     
     
     1
     
     
     
     ]);    

     
     
     
     18 
     
     
     
         system(
     
     
     
     "
     
     
     
     PAUSE
     
     
     
     "
     
     
     
     );

     
     
     
     19 
     
     
     
         
     
     
     
     return
     
     
     
      EXIT_SUCCESS;

     
     
     
     20 
     
     
     
     }

     
     
     
     21

COJ 1016 Simple calculations 难度在于推导求解a1的公式，编程很简单

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