354. Missax -

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

missing = 0 for i = 1 … N+1 missing ^= i repeat N times read x missing ^= x output missing We prove the sum‑based algorithm; the XOR version follows the same line of reasoning. Lemma 1 Let S = Σ_{i=1}^{N+1} i . Let T = Σ_{j=1}^{N} a_j be the sum of the numbers actually present. If exactly one element m of {1,…,N+1} is missing, then S - T = m .

N a1 a2 … aN (may be split over several lines) The file ends with a line containing 0 , which must be processed.