How to solve these kind of questions? I used BigInteger in java to find the product resulted in tle. Can someone help me or give a clue? Please, thanks in advance
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here answer is n%m always 
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How? Can you please explain?
easy buddy product of (a[i] * a[i+1] * a[i+2] ----- *a[n] )%a[i]=0
now in question we have to find sum for each i is (a[i]a[i+1]----- *a[n] +1 )%a[i] = (0+1)=1
now this 1 add upto n times
1+1+1+1------n times = n
now print n%m
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for ex : (5 * 4 * 6 * 3 )%5 =0 as one factor of 5 is present in multiplicants ie., 5
now (5* 4 * 6 * 3 +1 )%5 = 1%5 = 1
or (a+b)%m = (a%m+b%m)%m
(a-b)%m = (a%m-b%m + m)%m
(a*b)%m = (a%m * b%m)%m
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I don’t think this is correct.
1%1 is 0 and not 1
Damn bro… This is way basic math. Omg 
thank you so much. Can you please tell me this one too
[Programming Problems and Competitions :: HackerRank] . I did n/m and then checked divisibility by 2. I got WA. Thanks a lot bro
yes , so which part i said wrong ?
Ans is just
(sum (1\mod a[i]) )%m for 0 \le i \le n-1
Because,
(x*y*z) \mod x = 0
(x*y*z) \mod y = 0
(x*y*z) \mod z = 0
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if a[i] is 1 then its contribution will be zero right ?
because 1 modulo 1 is 0
1
5 2
1 1 1 1 1
Should be 0
Not 5%2
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Same here
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hell yes… i forgot 
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Awwwwwwww 
They have weak test cases 
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Why? Your logic have AC?
yesss
print(n%m)
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Damn… weak af… Can you please help me in this too… please please please
I divide the n/m and then checked whether it’s divisible by 2 or not. It passed half of the test cases… where must I have gone wrong?
Not weak, Wrong test cases LOL.
They also forgot it.
Test Case #3 fails with my soln
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bhai bhai