IT討論區(116) 加拿大救生艇門已開

freecodecamp 條條片都567粒鐘，點睇呀

lang嚟講有呢啲operator都算係基本要求

mask_11 應該點樣理解

Actually this can be even more simplified if we implement a 2-bit counter using boolean logic ourselves:

std::vector<int64_t> input;

// The bits in a & b implement 64 parallel 2-bit counters to keep track 
// of number of bits in input at position i. a[i] is bit 0 and b[i] bit 1 of the
// counter for position i.
int64_t a = 0, b = 0;

// After this loop finishes, a hold the number that appears once and
// b the number that appears twice.
for (int64_t n_i : input) {
  b ^= a & n_i;
  a ^= n_i;
}

I think this is more efficient and elegant.

我用4個bit黎儲一個2-bit number，加完1之後有機會overflow，咁and mask_11就會清除overflow bit，等結果係2個bit。其實用個兩個mask應該叫mask_0001同mask_0011會清楚啲。要防止overflow去隔離個counter其實三個bit都得但兩個bit mask唔會係1111 同 3333咁容易寫，人手有機會寫錯。

想問下
一般做trainee ot 冇補水 正常？

完全跟唔到大佬思路

突然唔想轉行
宜家份工就係正係唔洗ot
就算ot 都有補水

基本野...
你明唔明概念上要maintain bit counter先？

所以等於modulo 4？
omg

yes, mod 4 得 0, 1, 2 同 3 用兩個bit儲到。一個64-bit variable可以分成32個2-bit counter。但係咁樣做有一個問題係加數進位會overflow入隔離個counter，所以要3個bit先方便。每一次計完加數後mask off返個overflow bit就可以解決進位嘅問題。但3個bit又會令到人手寫bit mask constant教難，所以就用4個bit。

如果呢有睇過SIMD programming（MMX,SSE, AVX, AltiVec, VSX, Neon....) 其實概念上好類似，不過我同integer unit嘅ALU黎做parallel addition，要人手解決overflow嘅問題咁解。

係 就係mod4

IT9 OT冇補水係正常
正常計補水係啲weekend/平日半夜嘅official scheduled work就可能有得補

Official scheduled work分分鐘都無
好撚鐘意office hour 9-6 但schedule 7 8點做個task然後話第二朝一定要