Uvaoj 10014-simple calculations

I held a game on HUST and added some simple mathematical questions in Liu lujia's book.

Today, I am not going to go to the server. You still don't know what it looks like. Hust will return if the question is handed in.Judging Error 1.

Let's just explain this question first.

Question

This gives a formula A _{I =
( I-1 + I + 1)/2-C I  (1 <= I <= N ).
A is A 0 ,
A 1 ,..., A N + 1(N
<= 3000;-1000 <= I 1000) Yes Sequence of N + 2 Elements . Here you are. A 0 ,
A N + 1, C1,
..., CNYou need to findA1}

Derivation:

First, superimpose the N formulas that can be obtained according to the formula, obtain a [1] + A [n] = A [0] + A [n + 1]-2 (C [1] + .... + C [N]), transformed to a [0]-A [1]-2 (C [1] +... + C [N]) = A [n]-A [n + 1]. Therefore, we can represent any A [k]-A [k + 1] as a [0]-A [1] (because C is known ). Perform another superposition. The left side of the formula is (a [0]-A [1]) + (A [1]-A [2]) +... (A [n]-A [n + 1])
= A [0]-A [n + 1], used on the rightA [0]-A [1], and the right side is (a [0]-A [1]) + (A [0]-A [1]-2 sigma (C [K], k = 1 ~ 1) +... = (n + 1) A [0]-(n + 1) A [1]-2 (SIGMA (C [1 ~ 1]) + sigma (C [1 ~ 2]) +... + sigma (C [1 ~ N]). in this way, only the iterative superposition of the C array is required for the Sigma part. The formula for a [0] And a [n + 1] to obtain a [1] can be simplified.

Therefore, the idea is to create a stacked sub-operator to separate the two sides of the equation.

Code(Ac ):

# Include <cstdio> # include <cstring> # include <iostream> using namespace STD; # define maxn 3002 double C [maxn]; int N; double solve (double A0, double an1) {double T = 0; For (INT I = 1; I <= N; I ++) {C [I] + = C [I-1]; T + = C [I];} double ret = N * A0 + an1-2 * t; return ret/(double) (n + 1);} int main () {int t; scanf ("% d", & T); While (t --) {double A0, an1; scanf ("% d", & N ); scanf ("% lf", & A0, & an1); For (INT I = 1; I <= N; I ++) {scanf ("% lf ", & C [I]);} double ans = solve (A0, an1); printf ("%. 2lf \ n ", ANS); If (t) printf (" \ n ");}}