354. Missax Apr 2026

int main() { ios::sync_with_stdio(false); cin.tie(nullptr); long long N; while (cin >> N) { if (N == 0) break; // end of input // ----- sum based solution ----- long long missing = (N + 1) * (N + 2) / 2; // Σ_{i=1}^{N+1} i for (long long i = 0, x; i < N; ++i) { cin >> x; missing -= x; } cout << missing << '\n'; /* ----- xor based solution (alternatively) ----- long long missing = 0; for (long long i = 1; i <= N + 1; ++i) missing ^= i; for (long long i = 0, x; i < N; ++i) { cin >> x; missing ^= x; } cout << missing << '\n'; ------------------------------------------------- */ } return 0; } The program follows exactly the algorithm proved correct above, conforms to the required I/O format and runs in linear time with constant extra memory. It compiles under any standard C++17 compiler.

read N if N == 0 → finish missing = (N+1)*(N+2)/2 // 64‑bit integer repeat N times read x missing -= x output missing or (XOR version)

Proof. The algorithm first stores missing = S . During the input loop it subtracts each read number a_j from missing . After the loop finishes 354. Missax

S = (sum of present numbers) + m = T + m Rearranging gives m = S – T . ∎ The algorithm computes missing = S – T .

N a1 a2 … aN (may be split over several lines) The file ends with a line containing 0 , which must be processed. int main() { ios::sync_with_stdio(false); cin

Proof. All numbers of {1,…,N+1} appear either in T (if they are present) or are the missing value m . Hence

missing = S – Σ a_j = S – T ∎ For each test case the algorithm outputs the unique missing integer. The algorithm first stores missing = S

Proof. By Lemma 2 the value stored in missing after processing the whole test case equals S – T . By Lemma 1 S – T equals the missing element m . Therefore the printed value is exactly m . ∎ Time – each number is read and processed once → O(N) per test case. Memory – only a few 64‑bit variables are kept → O(1) . 6. Reference implementation (C++17) #include <bits/stdc++.h> using namespace std;

(Typical “find the missing element” problem – often appears on many online judges under the name Missax .) 1. Problem statement You are given an integer N ( 1 ≤ N ≤ 10⁶ ) . Then N distinct integers a₁ , a₂ , … , a_N are supplied.