Pretty annoying, not really a mathematics challenge, but rather a "know-how-to-deal-with-long-integers" challenge.

If you're failing all but 1 of the test cases, you more than likely did not implement the modulo described in the instructions. In my opinion this problem would be much better if we didn't have to handle data overflow in such a way. In any case it's only 1 extra line of code:

var routeCount=1;
for(int i=0;i<routes.Length;i++){
	routeCount*=routes[i];
	routeCount%=1234567;
}
return routeCount;

For C# LINQ

int result = arr.Aggregate(1, (a, b) => (b * a)%1234567);

Snippet.

for(int j = 0; j < towns - 1; j++) {
                routes *= scanner.nextInt();
                for(;;) {
                if (routes > 1234567) routes = routes - 1234567;
                    else break;
            	}
}

easy question

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        long long sum=1;
        for(int i=0;i<n-1;i++)
        {
            
            long long x;
            cin>>x;
            sum=((sum%1234567)*(x%1234567))%1234567;
            
        }
        cout<<sum<<endl;
    }
    return 0;
}

The modulo makes me laugh a bit... they have this story layer with Gandalf traveling between towns to make the question make sense. Then, with no explanation, the answer must be submitted as a modulo.

You weird, Gandalf.

Those were some biga$$ numbers.

while multiplying all the elements in a list and printing the output, how is

for i in list: ans *= i print (int(fmod(ans,1234567)))

and

print (functools.reduce(op.mul, s, 1) % 1234567)

different. As, in one case solution is being accepted while in another it just passes only one testcase.

def connectingTowns(n, routes):

prod = 1
for i in routes:
    prod *= i
return prod%1234567

Haskell implementation using a fold

solve :: [Int] -> Int
solve = foldl1 (\m n -> m * n `mod` 1234567)