There is a contest! | В настоящий момент идёт турнир. Некоторая информация и функции сервера недоступны до его окончания.
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| imesecan | Apr.08.2011 at 05:20:49 PM | |
| 0 | Post is deleted
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| imesecan | Apr.09.2011 at 10:19:41 AM | |
| 1 | Unlucky :)
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| imesecan | Apr.09.2011 at 09:56:40 PM | |
| 2 | I think you don't need reverse function. When calculating the remainders you can directly add this remainder to the beginning of resulting number:
You can use multiplier as the powers of 10, The first number is to be multiplied with. Let's say that the first modulus operation gives 2 and res is initially 0 and the number is 17
res = res + 1 * rem; ==> res= 2
then for the second remainder
res = res + 10 * rem; ==> res= 2 + 2* 10 ==> 22
then on the third step
res = res + 100 * rem; ==> 22 + 1 * 100 ==> 122
so on.
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| imesecan | Mar.19.2013 at 01:15:45 PM | |
| 3 | Post is deleted
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| imesecan | Mar.24.2013 at 10:46:50 AM | |
| 4 | Post is deleted
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