481 B
481 B
I like the occasionally really easy problems between the others. It's like a little break sometimes.
The "trick" here is to understand the modulo operator. If you have (a + b) % c
you can also write a%c + b%c
.
And also you can write (a * b)%c
as (a%c) * (b%c)
.
So all we do is calculate the sum kinda normally, but we do modulo 10^10
after each step (every addition and multiplication).
We guarantee this way that out numbers never exceed the range of an 64bit integer.