Opening#13087

Opening (Editorial)

Problem Author:	Hasan Saleh (MathModel)
Prepared by:	Hasan Saleh (MathModel)
Edited by:	Hasan Saleh (MathModel)

The problem can shortened to Given two integers $n$ and $m$ Give $a_{1}$ a_1 of $a_{1}, a_{2}, \dots, a_{n}$ a_1,a_2,\ldots,a_n for which

$a_{i + 1} - a_{i} = m$ a_{i+1}-a_i=m for all $(1 \leq i < n)$ (1 \le i <n)
$a_{n} - a_{1} = median$ a_n-a_1=\text{median}

Here , Notice that median is $\frac{n}{2}$ \frac{n}{2} when $n$ is even and $\frac{n + 1}{2}$ \frac{n+1}{2} when $n$ is odd , this is exactly how ceil function would work so the median always lies in index $⌈ \frac{n}{2} ⌉$ \lceil \frac{n}{2} \rceil.

$a_{i + 1} - a_{i} = m$ a_{i+1}-a_i=m for all $(1 \leq i < n)$ (1 \le i <n)
$a_{n} - a_{1} = a_{⌈ \frac{n}{2} ⌉}$ a_n-a_1=a_{\lceil \frac{n}{2} \rceil}

We've this fact :

a_{n} - a_{1} = a_{⌈ \frac{n}{2} ⌉}

a_n-a_1 = a_{\lceil \frac{n}{2} \rceil}

The General form of A.P. (Arithmetic Progression) is $a_{i} = a_{1} + m (i - 1)$ a_i=a_1+m(i-1) since each time we move we to add $m$ we get

(n - 1) m = a_{1} + (⌈ \frac{n}{2} ⌉ - 1) m

(n-1)m = a_1+(\lceil \frac{n}{2} \rceil-1)m

(n - 1) m = a_{1} + (⌈ \frac{n}{2} ⌉ - 1) m

(n-1)m = a_1+\boxed{(\lceil \frac{n}{2} \rceil-1)m}

Moving the boxed part to Left Hand Side (LHS)

(n - 1 - ⌈ \frac{n}{2} ⌉ + 1) m = a_{1}

(n-1-\lceil \frac{n}{2} \rceil+1)m = a_1

Recall the fact that $⌊ \frac{n}{2} ⌋ + ⌈ \frac{n}{2} ⌉ = n$ \lfloor \frac{n}{2} \rfloor+\lceil \frac{n}{2} \rceil=n for all integer $n$ thus $n - ⌈ \frac{n}{2} ⌉ = ⌊ \frac{n}{2} ⌋$ n-\lceil \frac{n}{2} \rceil=\lfloor \frac{n}{2} \rfloor

a_{1} = m \cdot ⌊ \frac{n}{2} ⌋

\boxed{a_1=m \cdot \lfloor \frac{n}{2} \rfloor}

Complexity is $O (1)$ .

def solve():
   n,m=map(int,input().split())
   print(m*(n//2))

for _ in range(int(input())):
   solve()