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- HAI 1.2
- I HAS A I обьявили переменную I
- I R MAEK I NUMBAR дали тип number
- HOW IZ I FOO YR I обьявление функции foo с одним аргументом I
- BOTH SAEM I AN SMALLR OF I AN 1, O RLY?
- YA RLY
- FOUND YR I вернуть I, если I меньше единицы
- NO WAI Иначе
- I HAS A SMTN ITZ I IZ FOO YR DIFF OF I AN 1 MKAY обьявление переменной SMTN = foo(I-1)
- I IZ FOO YR DIFF OF I AN 2 MKAY I = foo(I-2)
- SMTN R SUM OF IT AN SMTN SMTN = SMTN+I
- FOUND YR SMTN вернуть SMTN
- OIC
- IF U SAY SO конец функции foo
- I HAS A POWR ITZ 1 oбьявление POWR=1
- POWR R MAEK POWR A NUMBAR дали POWR тип number
- IM IN YR HER UPPIN YR B TIL DIFFRINT B AN SMALLR OF B AN 1000 цикл B=0 B++ <= 1000
- POWR R QUOSHUNT OF POWR AN 10.0 POWR /= 10
- I = foo(b)
- PRODUKT OF POWR AN I IZ FOO YR MAEK B A NUMBR MKAY IT = POWR * I
- I R SUM OF I AN IT I += IT
- IM OUTTA YR HER конец цикла
- VISIBLE I вывести I
- KTHXBYE
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