From Lemma 1 every increment corresponds to a genuine mutual‑love pair. From Lemma 2 every genuine pair contributes exactly one increment. From Lemma 3 no non‑mutual pair contributes any increment. Therefore the total number of increments equals precisely the number of mutual‑love pairs. ∎ 5️⃣ Complexity analysis Time – The loop visits each of the N people once, performing O(1) work per iteration: O(N) per test case.
def solve() -> None: data = sys.stdin.buffer.read().split() it = iter(data) t = int(next(it)) out_lines = [] for _ in range(t): n = int(next(it)) love = [0] + [int(next(it)) for _ in range(n)] # 1‑based list ans = 0 for i in range(1, n + 1): j = love[i] if i < j and love[j] == i: ans += 1 out_lines.append(str(ans)) sys.stdout.write("\n".join(out_lines)) 412. Sislovesme
love[1 … N] // 1‑based indexing where love[i] = j means person i loves person j . From Lemma 1 every increment corresponds to a
Multiple test cases are given. T // number of test cases (1 ≤ T ≤ 20) N // number of people (1 ≤ N ≤ 10^5) love[1] love[2] … love[N] // N integers, 1 ≤ love[i] ≤ N The sum of N over all test cases does not exceed 10^6 . Output For each test case output a single line containing the number of mutual‑love pairs. Sample Input Therefore the total number of increments equals precisely
2 4 2 1 4 3 5 2 3 1 5 4
(A classic “mutual‑love” counting problem – often seen on SPOJ, LightOJ, and other online judges) 1️⃣ Problem statement You are given a group of N people, numbered from 1 to N . Each person loves exactly one other person (possibly himself). The love‑relationships are described by an array
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