Given two matrices A and B, both of them having N rows and M columns. The cells of A are numbered in row major order, and cells of B are numbered in column major order. How many cells appears at the same location in both the orderings ?
Explanation:
For 0 ≤ i < N, 0 ≤ j < M, the (i,j)th entries are:
Ai, j = i * M + j + 1
Bi, j = j * N + i + 1
equating the two, we get
i * M + j + 1 = j * N + i + 1
⇒ i * (M-1) = j * (N-1)
⇒ i / j = (N-1) / (M-1) = p / q
⇒ i = l * p, j = l * q for 0 ≤ l ≤ min((N-1) / p, (M-1)/q)
where p/q is reduced form of (N-1)/(M-1).
However, if g = gcd(N-1, M-1),
(p, q) = ((N-1)/g, (M-1)/g)
Therefore, (N-1) / p = (M-1) / q = g
and the cells which get same numbering are (l * p, l * q) for 0 ≤ l ≤ g, which are g+1 in number.
Boundary cases
N = 1 or M = 1, in which case the answer is max(N, M).
actually there is no need of boundary conditions.
in case when (n==1||m==1) then gcd(0,n(or)m) will be 0 and the answer will itself become 1 and correct.
enter code here
static void Main(string[] args)
{
int T = int.Parse(Console.ReadLine());
int cnt2;
for (int t = 0; t < T; t++)
{
string[] s = Console.ReadLine().Split();
int n = int.Parse(s[0]);
int m = int.Parse(s[1]);
if (n == 1 || m == 1)
cnt2 = Math.Max(n, m);
else
cnt2 = GCD(n-1, m-1) + 1;
Console.WriteLine(cnt2);
}
}
static int GCD(int a, int b)
{
if (b == 0) return a;
return GCD(b, a % b);
}
long long gcd(long long a,long long b)
{while(a>0&&b>0)
if a>b a%=b;
else b%=a;
return a+b;
}
so if either m=1 or n=1 then g=gcd(m-1,n-1) becomes zero and g+1 will give 1 which is correct answer!!
Manal, I think you mean “if (n==1 && m==1)” then output should be 1. Otherwise, if n==1 or m==1 then the answer would be the other number (e.g. n=1, m=5, then output 5).
@utkarsh_lath [Not something relating to this problem] I remember a problem from some previous contest, where 0^0 was conveniently taken as 1. That didn’t go by strict definitions.
Its all about conventions and widely accepted norms. Whether 0^0 is 0 or 1 is usually clear from the context.
Sometimes people give special mention to such things, sometimes assume that others are aware of it.