之前聽過 aws swe iii offer base 300k around
但其實唔覺得吸引
IT討論區(82) IT狗對住新老細:肅立起敬頌讚祂!!表示尊敬跪拜祂!!
素睛
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sde iii / manager
其實兩者差唔多
其實兩者差唔多
No you dun count words using r*nCr, each word has different length
You count total character used by a word by length*appearance
r*nCr is for comma between words
You count total character used by a word by length*appearance
r*nCr is for comma between words
係囉呢個係team lead level
好難外面入通常要入面升，仲要9 成人都上唔到
好難外面入通常要入面升，仲要9 成人都上唔到
300K好似交1/3稅，咁計返都有11萬港紙一個月喎
你諗下每個word會總共出現幾多次
另外r*nCr好易計的 你可以直接爆下 又可以d下(1+x)^n
另外r*nCr好易計的 你可以直接爆下 又可以d下(1+x)^n
供樓 + prop tax. 至少5k USD per month 好似
做左一年software engineer trainee
終於升職....
升撚左做senior software engineer trainee
終於升職....
升撚左做senior software engineer trainee
幾好笑 7/10
死唔少人
出年升做chief software engineer trainee
係
唔好成日屌我問埋啲四種fibonacci既寫法 因為我真係可以篩走連naive都唔識既on9仔
唔好成日屌我問埋啲四種fibonacci既寫法 因為我真係可以篩走連naive都唔識既on9仔
漏左software analyst trainee 同software manager trainee
Ng sick Sor lo
考吓fizzbuzz咪算
典解要蝦禾吾sick數學
典解要蝦禾吾sick數學
junior software engineer trainee leader
佢咁樣有排升 LOLZ
因為可以一題分到candidate係唔識programming 定係識programming但唔識maths 定係又識pgrogramming又識少少applied linear algebra 定係又識programming又識少少difference equation
比多幾級你升，等你加多幾次人工，仲唔跪謝腦細
一級加五百
講明係垃圾
sorry, but that one is also simple:
nCr = nC(n-r)
sum_{r = 0 .. n} r * nCr
= sum_{r= 0...n} (n-r) * nC(n-r) // sum in reverse order
=sum_{r = 0..n} (n-r) * nCr // nCr = nC(n-r)
2 * sum_{r = 0..n} r * nCr
= sum_{r=0..n} r * nCr + sum_r{r=0..n} (n-r)*nCr
= sum_{r=0..n} (r + n-r) * nCr
= n * sum{r=0..n} nCr
= n * 2^n
so sum{0..n} nCr = n/2 * 2^n
Intuitively, every subset of an n-element set correspond to an integer in [0..2^n] such that the i-th element is in the subset if the bit i is set. So we are really counting the number of set bits from [0..2^n]. Well, there are n * 2^n bits total (either 0 or 1) by symmetry only half of the bits are set so the number of 1's is n/2 * 2^n.
nCr = nC(n-r)
sum_{r = 0 .. n} r * nCr
= sum_{r= 0...n} (n-r) * nC(n-r) // sum in reverse order
=sum_{r = 0..n} (n-r) * nCr // nCr = nC(n-r)
2 * sum_{r = 0..n} r * nCr
= sum_{r=0..n} r * nCr + sum_r{r=0..n} (n-r)*nCr
= sum_{r=0..n} (r + n-r) * nCr
= n * sum{r=0..n} nCr
= n * 2^n
so sum{0..n} nCr = n/2 * 2^n
Intuitively, every subset of an n-element set correspond to an integer in [0..2^n] such that the i-th element is in the subset if the bit i is set. So we are really counting the number of set bits from [0..2^n]. Well, there are n * 2^n bits total (either 0 or 1) by symmetry only half of the bits are set so the number of 1's is n/2 * 2^n.