# I just solved this coding challenge, but I don't understand why my solution works 🤔

### Kinyanjui Wangonya ・1 min read

I've been trying to do one programming challenge a day on Kattis, and I just solved this one. I really had no idea how to solve it at first, so I just played around with the sample input/output data provided and noticed a pattern:

In the final sample, an input of 10 and 10 gives an output of 91: that's `10 * (10-1) + 1)`

. Taking the first input to be `x`

and the second to be `y`

, this gives a formula of `x * (y-1) + 1`

which gives the correct output for all the other inputs and passes all test cases:

```
# https://open.kattis.com/problems/faktor
import sys
def faktor(articles, impact):
print(int(articles)*(int(impact)-1) + 1)
if __name__ == '__main__':
a, i = sys.stdin.readline().split()
faktor(a, i)
```

The thing is, that formula doesn't seem to have anything to do with the question in the challenge. Maybe I'm missing something 🤔

Actually, your formula is correct. Let's derive it mathematically.

The number of articles you want to publish is

`A`

, and the impact you want is`I`

. It's given that`impact(I) = total_citations(C) / total_articles(A)`

. It's also given thatrounding is done only upwards.For example if your impact value is`23.0000001`

, then it will be rounded up as`24`

. (That's same as saying rounding is done using`math.ceil`

).We want to find the minimum number of citations

`C_min`

so that our rounded value of impact equals to`I`

. We know that`I = math.ceil( C / A )`

, therefore, the value of`(C / A)`

must be something between`I-1`

and`I`

. That is,`I-1 < (C/A) <= I`

. From this, we can write,`(I-1)*A < C <= I*A`

.Now we know that value of

`C`

should be strictly higher than`(I-1)*A`

and it should be less than or equal to`I*A`

.`C`

is an integer, therefore the minimum value of`C`

that satisfies the above condition must be`(I-1)*A + 1`

.Hence it's proved that,

`C_min = (I-1)*A + 1`

.Wow, it all makes sense now. Thanks a lot @gnsp

Quick! Take your good luck and pass a coding challenge at a financial institution like Charles Schwab. They'll never see you coming and won't notice your paycheck once you're there!

A*I-(A-1)