UV-11774 Doom's Day (regular)

 

We all know about the legend oftower of Hanoi. it is said that the world will end after finishing the puzzle. what we don't know is another legend about when the world will end which is verifiedby the scientists.

 

It is all about a 3 ^ N * 3 ^ m grid. initially the grid is filled with 1 to 3 ^ (m + n) in row major order. at each stepthe puzzle is rearranged by reading it in row major order and putting them incollumn major order. see the following examples.

 

1	2	3	To	1	10	19
4	5	6		2	11	20
7	8	9		3	12	21
10	11	12		4	13	22
13	14	15		5	14	23
16	17	18		6	15	24
19	20	21		7	16	25
22	23	24		8	17	26
25	26	27		9	18	27

 

 

1	2	3	To	1	4	7
4	5	6		2	5	8
7	8	9		3	6	9

 

Now every day the puzzle is rearrangedonce. The legend says if someday initial configuration returns the world willend. Now you are wondering when the world is going to end.

 

Input

Input starts with a linecontaining Number of test casesT ≤ 10000. Each test case containstwo positive integerM ≤ 10 ^ 9AndN ≤ 10 ^ 9.

 

Output

For each case print one linecontaining days before dooms day. The input will be such that this number fitsin 64 bit unsigned integer.

 

Sampleinput outputfor sample input

5 1 1 1 2 3 1 2 2 98876767 12234

Case 1: 2 Case 2: 3 Case 3: 4 Case 4: 2 Case 5: 98889001

 

Problemsetter:Tanaeem Md. Moosa

Specialthanks:Jane Alam Jan

To the number of 1 to 3 ^ (n + M) in the 3 ^ N * 3 ^ m lattice, each operation is to prioritize the number by row, put it in by column first, and return to the initial state after how many times

Idea: Regular ans = (a + B)/gcd (A, B );

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}int main() {int t, a, b, cas = 1;scanf("%d", &t);while (t--) {scanf("%d%d", &a, &b);printf("Case %d: %d\n", cas++, (a+b)/gcd(a, b));}return 0;}

UV-11774 Doom's Day (regular)