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23.12.2017, 21:09 | #1 |
Регистрация: 23.12.2017
Сообщений: 3
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Задача на нахождение кол-ва делителей числа a
Решаю задачу на нахождение кол-ва делителей с сайта http://informatics.mccme.ru:
var a,b,i:integer;k:word; begin readln(a); if a>=1 then begin for i:=1 to a div 2 do if a mod i=0 then b:=b+1; k:=b+1; writeln(k); end; end. При отправке решение не засчитывается , выдаёт: "Hеправильный формат вывода". Помогите поскорее Заранее спасибо! |
23.12.2017, 21:25 | #2 |
Старожил
Регистрация: 17.11.2010
Сообщений: 19,042
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И что, искать на том сайте в каком формате вывод должен быть? Да еще поскорее, юморист ))
Если бы архитекторы строили здания так, как программисты пишут программы, то первый залетевший дятел разрушил бы цивилизацию
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23.12.2017, 21:27 | #3 |
Регистрация: 23.12.2017
Сообщений: 3
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Вот))
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04.02.2018, 20:59 | #4 |
Регистрация: 23.12.2017
Сообщений: 3
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ХАХАХАХАХАХАХАХАХАХАХАХАХАХАХАХАХАХ АХА
НУ ВОТ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!! var x,l,i,k:longint;koren:real; begin readln(x); koren:=sqrt(x); l:=trunc(koren); for i:=1 to l do if x mod i=0 then k:=k+1; k:=k*2; if frac(koren)=0 then writeln(k-1) else writeln(k); end. |
10.02.2018, 08:30 | #5 |
Я самый любопытный
Участник клуба
Регистрация: 24.07.2012
Сообщений: 1,949
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Проверил 1-й пост. У меня работает.
Может подправить код: Код:
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