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;

Snippet.

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

For C# LINQ

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

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;
}

Those were some biga$$ numbers.

For C++

Perform mod after every multiplication or you'll exceed size limit

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.

This is very good, it will be on my saved recipes list.

driving directions

I'm not able to understand why are we using modulo 1234567?

Please answer me. How to do the problem with C-Programming?