stzsch Pretty annoying, not really a mathematics challenge, but rather a "know-how-to-deal-with-long-integers" challenge.
mike006322 The way modular arithmatic works, you can take the mod of each component of the product before you calculate the product and get the same result as if you were to mod the poduct after multiplying. My code had something like:
int product = 1 for (i<n-1) { product = product*T[i]%1234567 }
jeremydcarr Thanks for that explanation. I was wondering why my solution of doing it after the multiplication was not working. I figured it was overflow causing issues, but I didn't think to solve that like this.
jongod5399 [deleted]
maheshvangala191 why 1234567 why not the other number be taken?
mike006322 "Output Format Total number of routes from T1 to Tn modulo 1234567"
The problem says to output mod 1234567. That is where that number comes from.
piyush1751995 shouldn't there be product = product%123456 after the loop.
I have studied (A * B) mod C = (A mod C * B mod C) mod C
cfranco 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;
nilkun Snippet.
for(int j = 0; j < towns - 1; j++) { routes *= scanner.nextInt(); for(;;) { if (routes > 1234567) routes = routes - 1234567; else break; } }
Marcinho Awesome solution! Thank you :)
piyush1751995 take remainder directly
ROCKET_RONALDO 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; }
gbekosa Why are you using mod 1234567 3 times? You only need to use it once after multiplying sum by x.
sum = (sum * x) % 1234567;
Learner_College BCoz it is a modulo property. (a*b)%c=((a%c)*(b%c))%c
AdarkTheCodr That is the property, yes, but you can remove the first two modulos because a)
xis between 1 and 1000, sox modulo 1234567is pointless, and b) modulo-ingsumbefore it is used in the equation and before assigning the result back tosumis redundant.
c650Alpha Those were some biga$$ numbers.
driftwood For C++
Perform mod after every multiplication or you'll exceed size limit
ankitarun 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.
manvee_2603 why we are using mod1234567
RSTHW read format output again
manvee_2603 why we are using mod1234567
johnlr fmodreturns a float, so you will print extra.0on each termquick_dudley Forgetting to output mod some number is the #1 mistake I make on hackerrank problems.
itsbaldeep JavaScript
Just use Array.reduce() and keep multiplying
accumulator (a) by element (b) modulo 1234567
Note: Set initial value to 1, by giving it a second argument
routes.reduce((a, b) => a * b % 1234567, 1);
kvsprasadu My solution in java,
static long connectingTowns(int n, int[] routes) { return Arrays.stream(routes).boxed().reduce((x,y) -> (x % 1234567) * (y % 1234567)).orElse(0) % 1234567; }
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