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copied!<p>You have 4 rules laid out so just write code to check each rule.</p> <ul> <li><p>The number is N digits. Determine what N is. Could be done by converting to a string or a vector for each digit. You may need to use division by 10 and mod (%) 10 a few times.</p></li> <li><p>A method to move from one digit to the next. So if you are at digit position x with value y, you move to x+y mod N that is (x+y)%N. The first position is considered position 0, of course.</p></li> <li><p>You need to check that you touch every position, and also that you do not have a duplicate. This could be one check. If you reach a digit you have seen before you know you have the correct solution if and only if this is the Nth iteration and you are at index 0.</p></li> </ul> <p>12 is failure. Because although it is after 2 iterations you do not reach index 0 but revisit index 1 (2 takes you from index 1 back to index 1)</p> <p>11 is failure because you see another 1 when you have not had all your iterations yet.</p> <p>123 is failure because you get back to 1 an iteration too early. You don't need to know you didn't see the 3, or that you got back to index 0, just that you saw another 1 too early.</p> <p>285 works though. You see 2 then 5 then 8 then 2. You are at index 0 and have had 3 iterations.</p> <p>You need to store a collection of seen digits. vector is slightly controversial, so you could use std::bitset or even just a bool[10] will do for this example, or vector. You could also use std::set, the latter case of which would allow you to check its size to see your iteration count.</p>

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