How’s that? from itertools import groupby input_list = [0, 1, 0, 0, 1, 1, 1, 0, 1, 0] results = [0, 0] # First value is zeros, second is ones for key, values in groupby(input_list): results[key] += (2**len(tuple(values))) – 1 assert results == [6,9] solved a pythonic way of doubling consecutive binary bits [closed]
That code isolates the two bits indicated by the arrows below, and moves them to the two least significant bit positions. vv 11111111111111111111111111111111 original value vv 00000000000000000011111111111111 after >>18 (shift right 18 positions) vv 00000000000000000000000000000011 after & 3 (mask out all but the 0th and 1st bits) This assumes an unsigned value, and no special … Read more
The following only assumes 2’s complement with at least 16 bits: int mask = ~0x7FFF; return !(x&mask)|!(~x&mask); That uses a 15-bit constant; if that is too big, you can construct it from three smaller constants, but that will push it over the 8-operator limit. An equivalent way of writing that is: int m = 0x7FFF; … Read more
For 16 bits, there are 216 = 65,536 possible settings. Each setting of bits can represent at most one value. So 16 bits of data can represent only 65,536 values. 32 bits have 232 = 4,294,967,296 possible settings. If your 32-bit data uses more than 65,536 of the possible values that could be represented, it … Read more
for(int i=0; i<(1<<N); i++) is used to do the combination: it loops from 0b00000 to 0b11111 The following code: long long sum=0; for(int j=0; j<N; j++) if(i & (1<<j)) sum+=a[j]; } does the sum according to the previous ‘mask’: so with i == 0b1010, sum will be a[1] + a[3] (mask index starts from right). … Read more
When n++ & 1 executes, lets expect n is odd number at the moment, condition is true. Post incrementation is applied and n is even when result += n executes. Thats why its counting even numbers instead of odd. Modify code to while(n < 99) result += !(n++ & 1) ? n : 0; and … Read more
The problem is here: if(0 == maps&(1<<k)){ The equality operator == has higher precedence than the bitwise-AND operator &. So the above evaluates to this: if ((0 == maps) & (1 << k )) { You need to add parenthesis to get the desired behavior: if (0 == (maps & (1 << k))) { You’ll … Read more
You can convert a number to hex like this: 0x8080.ToString(“x6”) The 6 is the number of decimal places. See the documentation for more detail. 2 solved how to convert 0x8080 to 8080 int? [closed]