You are given two arrays of integers a, b both with length n <= 10^5. For each 0 <= x <= n print the sum of a[i]*b[x-i] for all 0 <= i <= x.
It's obvious this can be done in quadratic time, but can we do any better?
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
You are given two arrays of integers a, b both with length n <= 10^5. For each 0 <= x <= n print the sum of a[i]*b[x-i] for all 0 <= i <= x.
It's obvious this can be done in quadratic time, but can we do any better?
Name |
---|
If P(x) = a0 + a1*x + ... + an * x^n and Q(x) = b0 + b1*x + ... + bn*x^n then how can you obtain the requested sum? How can you compute it?
Let R(x) = P(x) * Q(x). Your sum is R(1).
FFT
Thanks for the reply. Your strategy is quite interesting, however that only solves the problem if we want the sum over a[i]*b[x-i] for all x and for all i. The problem was to isolate each x and calculate the sum over a[i]*b[x-i] for all i for this value of x only, then move on to the next value of x. I'm very sorry if this wasn't clear.
Oh, I didn't get that. I think that the same strategy works in this case though.
It seems that the answer will always be equal to . (can't prove it, sorry)
Let p be the partial sum array of a. Then the answer is simply .