Идёт турнир! | В настоящий момент идёт турнир. Некоторая информация и функции сервера недоступны до его окончания.
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| imesecan | 01.окт.2015 в 10:49:59 | |
| 0 | -- Test 1 ---
17 8
--- Pattern ---
21
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| imesecan | 01.окт.2015 в 11:15:55 | |
| 1 | After 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 | 01.окт.2015 в 11:16:06 | |
| 2 | You don't have to specify in you solutions the limits, for example don't write: if (number >=0 && number <100000) . It is given in the question just to clarify you for the limits of the numbers. And try using long long int.
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