ABC177C

from ABC177 ABC177C C - Sum of product of pairs

• Take everything out and square it, then draw a diagonal line and divide it in half.
• 
• If the value you want is S $2S+\sum A_i^2=(\sum A_i{)}^2$ # Distribution law Half of the queue Exchange of product and sum.
• WA for naively truncating “divide by two” to a divisor.
  • If odd, add mods to make it an even number, then divide by 2 python

def solve(N, AS):
    sum = 0
    sumSq = 0
    for i in range(N):
        sum += AS[i]
        sum %= MOD
        sumSq += AS[i] * AS[i]
        sumSq %= MOD

    ret = (sum * sum - sumSq) % MOD
    if ret % 2 == 0:
        return ret // 2
    else:
        return (ret + MOD) // 2

• The official explanation is cumulative sum, which is a way to multiply a row by a single multiplication.
  • They said my solution was “not simply divisible by two, so it would be a difficult solution using the inverse original.”
    • It’s just too abstract and difficult to think about.
  • If you’ve ever tried it with a number as small as 11, it’s not hard to come up with. python

xs = [None] * 11
for i in range(11):
    xs[i * 2 % 11] = i
# xs => [0, 6, 1, 7, 2, 8, 3, 9, 4, 10, 5]

- There's an i in the second i like this.
- The odd-numbered number gets a value in the second round once you get to the end.

• 
  • Hmmm, I guess it was 4ms short of the fastest!
  • The fastest code takes the surplus only at the end.
    • Is it faster to push through with long integers if the maximum is about 10^30?
      • Fast long integers

---

This page is auto-translated from /nishio/ABC177C using DeepL. If you looks something interesting but the auto-translated English is not good enough to understand it, feel free to let me know at @nishio_en. I’m very happy to spread my thought to non-Japanese readers.