Ultraviolet A 10014-simple calculations

Simple calculations

The Problem

There is a sequence of N + 2 elements A0, A1 ,..., An + 1 (n <= 3000;-1000 <= ai 1000). It is known that
Ai = (ai-1 + AI + 1)/2-Ci
For each I = 1, 2,..., n. You are given A0, An + 1, C1,..., CN. Write a program which calculates A1.

The input

The first line is the number of test cases, followed by a blank line.

For each test case, the first line of an input file contains an integer n. the next two lines consist of numbers A0 and an + 1 each having two digits after decimal point, and the next n lines contain numbers Ci (also with two
Digits after decimal point), one number per line.

Each test case will be separated by a single line.

The output

For each test case, the output file shoshould contain A1 in the same format as A0 and an + 1.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

1150.5025.5010.15

Sample output

27.85

A [I] = (a [I-1] + A [I + 1])/2-C [I];
2 * (a [I] + C [I]) = A [I-1] + A [I + 1]

A [1]-A [0] = A [2]-A [1]-2 * C [1]
A [2]-A [1] = A [3]-A [2]-2 * C [2]
...
A [I]-A [I-1] = A [I + 1]-A [I]-2 * C [I]
A [I]-A [0] = A [I + 1]-A [1]-2C

A [1] = A [I + 1]-A [I] + A [0]-2C (n) [n items and]
Do this no idea, it is estimated that most people and I Y-Y, as long as the accumulation of the answer will come out.
Success means sticking to what has already been done. O (success _ success) O Haha ~
N = 1 A [1] = A [2]-A [1] + A [0]-2C (1)
N = 2 A [1] = A [3]-A [2] + A [0]-2C (2)
N = 3 A [1] = A [4]-A [3] + A [0]-2C (3)
N * A [1] = A [n + 1]-A [1] + N * A [0]-2 * [C (1) +... C (n)]

(N + 1) A [1] = A [n + 1] + N * A [0]-2 * [C (1) + C (n)]

 

# Include <stdio. h>
# Include <math. h>
Void main ()
{Double C [3001], s [3001], X, Y;
Int N, I, T;
Scanf ("% d", & T );
While (t --)
{
Scanf ("% d", & N );
Scanf ("% lf", & X, & Y );
For (I = 1; I <= N; I ++)
Scanf ("% lf", & C [I]);
S [1] = C [1]; s [0] = s [1];
For (I = 2; I <= N; I ++)
{S [I] = s [I-1] + C [I];
S [0] + = s [I];
}
Printf ("%. 2lf \ n", (N * x + y-2 * s [0])/(n + 1 ));
If (t) printf ("\ n ");
}
}