Strange Table (Codeforces 1506A)
Tags: Math, math

Polycarp found a rectangular table consisting of $$$n$$$ rows and $$$m$$$ columns. He noticed that each cell of the table has its number, obtained by the following algorithm "by columns":   cells are numbered starting from one;  cells are numbered from left to right by columns, and inside each column from top to bottom;  number of each cell is an integer one greater than in the previous cell. For example, if $$$n = 3$$$ and $$$m = 5$$$, the table will be numbered as follows:$$$$$$ \begin{matrix} 1 & 4 & 7 & 10 & 13 \\ 2 & 5 & 8 & 11 & 14 \\ 3 & 6 & 9 & 12 & 15 \\ \end{matrix} $$$$$$However, Polycarp considers such numbering inconvenient. He likes the numbering "by rows":   cells are numbered starting from one;  cells are numbered from top to bottom by rows, and inside each row from left to right;  number of each cell is an integer one greater than the number of the previous cell. For example, if $$$n = 3$$$ and $$$m = 5$$$, then Polycarp likes the following table numbering: $$$$$$ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ \end{matrix} $$$$$$Polycarp doesn't have much time, so he asks you to find out what would be the cell number in the numbering "by rows", if in the numbering "by columns" the cell has the number $$$x$$$?

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$). Then $$$t$$$ test cases follow.Each test case consists of a single line containing three integers $$$n$$$, $$$m$$$, $$$x$$$ ($$$1 \le n, m \le 10^6$$$, $$$1 \le x \le n \cdot m$$$), where $$$n$$$ and $$$m$$$ are the number of rows and columns in the table, and $$$x$$$ is the cell number.Note that the numbers in some test cases do not fit into the $$$32$$$-bit integer type, so you must use at least the $$$64$$$-bit integer type of your programming language.

Output: For each test case, output the cell number in the numbering "by rows".


Sample Inputs:
5
1 1 1
2 2 3
3 5 11
100 100 7312
1000000 1000000 1000000000000

Sample Outputs:
1
2
9
1174
1000000000000

Solution (Python):
for a in [*open(0)][1:]:
    (b, c, d) = map(int, a.split())
    d = d - 1
    print(d // b + 1 + d % b * c)

Solution (Python):
for nik in [*open(0)][1:]:
    (c, n, v) = map(int, nik.split())
    v -= 1
    print(v % c * n + v // c + 1)

Solution (Python):
for s in [*open(0)][1:]:
    (n, m, x) = map(int, s.split())
    x -= 1
    print(x % n * m + x // n + 1)