Блог пользователя sharmaanuj334

Автор sharmaanuj334, история, 2 года назад, По-английски

I was practicing DP through CSES when I got to the Two Sets II problem, wherein I got the confusion on the usage of mod to get the correct answer.

I had read earlier that if we wish to take mod of a summation, we have to take mod of individual elements of the series and then take the mod of summation.

$$$ \text{ans} = ( \sum_{i=1}^{n} (a[i] \% \text{mod})) \% \text{mod} $$$

But here are my two submitted codes:

  1. Accepted: not taking the overall mod on summation.
  2. Wrong Answer: taking the overall mod on summation.
Wrong Answer Code
Accepted Code

Due to this, I got confused about how we should be taking the mod and, in general, how you tackle such situations. (Right now, I'm looking for a discussion like the one from which I got to know of the binary search method of while (r-l > 1), which I now use, so I can only stress more on the concept of the problem.)

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Whenever the sum can be bigger than the mod you must take a mod, which is almost always. For example if you add two numbers the max of each of them is MOD-1 so the sum can be 2*MOD-2. In CP, you don't need to decide on where to add mod though just put it after every operation. Some people use auxillary functions or even mod int templates.

PS: If the thing you are taking mod of can be negative you must make it positive because in C++ mod works in a way different than math (see Auxillary for example).

Auxillary
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CSES submissions are private.

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another thing — you can't just divide a number under mod to get the correct answer. you should be multiplying by the modular inverse.

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Modulo