There is a contest! | В настоящий момент идёт турнир. Некоторая информация и функции сервера недоступны до его окончания.
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| imesecan | Sep.29.2015 at 09:24:48 PM | |
| 0 | -- Test 1 ---
2 10
--- Pattern ---
4
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| imesecan | Dec.08.2015 at 11:12:53 AM | |
| 1 | It's faster if you check until the square root of the number.
while(divisor*divisor<=num)
You can also make faster than that.
if the number is 2, it's prime,
if it's not 2 and if the number is divisible by two, it's not prime,
otherwise start from 3 and check for only odd numbers until the square root of the numbers.
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