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    <p><code>secret</code> is always between 0 and 10,000,000 inclusive due to the mod by 10,000,001. The observed values are always the last 3 digits of <code>secret</code> (with leading zeros stripped) due to the mod by 1,000. So it's the other digits that are unknown and that only leaves us with 10,001 numbers to iterate over.</p> <p>For each <code>prefix</code> in 0..10,000, we start with <code>secret</code> being constructed from the digits of <code>prefix</code> followed by the first number in the observed sequence with leading zeros. If the list of generated numbers equals the observed list, we have a potential seed. If we end with no potential seeds, we know this must be a wrong machine. If we end with more than one, we don't have enough observations. Otherwise, we generate the next 10 values using the single seed.</p> <p>This runs in O(10,000NT), which is O(20,000,000) for the constraints given. Here's an extract from my solution in C++ (excuse the heavy use of macros, I only ever use them in competitions):</p> <pre><code>int N; cin &gt;&gt; N; int n[N]; REP(i, N) cin &gt;&gt; n[i]; ll poss = 0, seed = -1; FOR(prefix, 0, 10001) { ll num = prefix * 1000 + n[0]; bool ok = true; FOR(i, 1, N) { ll next = getRandomNumber(num); if (next != n[i]) { ok = false; break; } } if (ok) { poss++; seed = prefix * 1000 + n[0]; } } if (poss == 0) { cout &lt;&lt; "Wrong machine" &lt;&lt; endl; } else if (poss &gt; 1) { cout &lt;&lt; "Not enough observations" &lt;&lt; endl; } else { ll num = seed; FOR(i, 1, N) getRandomNumber(num); REP(i, 10) cout &lt;&lt; getRandomNumber(num) &lt;&lt; " "; cout &lt;&lt; endl; } </code></pre>
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    1. USmarcog
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    1. COSo you don't try to compute the initial secret. But how can you be sure the candidates you generate for the first observation are at all possible outputs? Can you show every number in [0, 10,000,000] has an invert with respect to `secret = (secret * 5402147 + 54321) % 10000001`?
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    2. CO@Wei Some won't be, but that case is already handled with the "Wrong machine" logic.
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    3. COOK, for any number o, there's always another number s, such that `o = (s * 5402147 + 54321) % 10000001`. s can be computed with `s = ((o - 54321) * 5194600) % 10000001`. So your algorithm is fine.
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