412. Sislovesme -

(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

When the loop later reaches i = b , the first condition fails ( b < a is false), so the pair is counted again. ∎ Lemma 3 If a pair i, j is not a mutual‑love pair, the algorithm never increments mutualPairs for it. 412. Sislovesme

long long ans = 0; // up to N/2 fits in int, but long long is safe for (int i = 1; i <= N; ++i) int j = love[i]; if (i < j && love[j] == i) ++ans; // count each 2‑cycle once cout << ans << '\n'; return 0; ∎ Lemma 3 If a pair i, j

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. From Lemma 3 no non‑mutual pair contributes any increment

love[1 … N] // 1‑based indexing where love[i] = j means person i loves person j .

import sys

Both limits satisfy the given constraints ( ∑ N ≤ 10⁶ ). Below are clean, production‑ready solutions in C++ (17) and Python 3 . Both follow the algorithm described above and use fast I/O to handle the maximum input size. C++ (GNU‑C++17) #include <bits/stdc++.h> using namespace std;