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Ali Akbar Tehsildar
a.tehsildar@gmail.com
32767 is the maximum value that can be stored in two bytes.
If using the compressed format, where a byte holds two decimal digits (because only 4 bits are needed to make nine), so two bytes would be four decimal digits, the largest which is 9999.
11b which is 1*2 + 1*1 = 3 would be for two bits. But a byte is 8 bits, so 2 bytes is 16 bits. The largest binary number is [2^16 - 1], which is 65535 (base ten)
To compute the largest value in an array, assume that the first element is the largest and store the value. Then traverse the remainder of the array. Each time a larger value is encountered, update the stored value. Once all values are traversed, return the stored value. In pseudocode, this algorithm would be implemented as follows: Algorithm: largest Input: array A of length N Output: largest value in A let largest = A[0] // store first value for index = 1 to N-1 // traverse remaining elements if A[index] > largest then largest = A[index] // update stored value if current value is larger next index return largest To determine the position of the largest value, we alter the algorithm as follows: Algorithm: largest_by_index Input: array A of length N Output: index of the largest value in A let largest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] > A[largest] then largest = index // update stored index next index return largest We can do the same to find the position of the smallest element: Algorithm: smallest_by_index Input: array A of length N Output: index of the smallest value in A let smallest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] < A[smallest] then smallest = index // update stored index next index return smallest To perform both algorithms simultaneously, we need to return two values. To achieve this we can use a simple data structure known as a pair: struct pair { int smallest; int largest; }; Algorithm: range_by_index Input: array A of length N Output: a pair indicating the position of the smallest and largest values in A pair result = {0, 0} // initialise the pair for index = 1 to N-1 // traverse remaining elements if A[index] < A[result.smallest] then result.smallest = index // update stored index else if A[index] > A[result.largest] then result.largest = index // update stored index next index return result
two thousand bits No, there are 8 bits in a byte.
0.0009765625 GB
It is estimated that each letter of a word is about two bytes. For example, a four-letter word would be around eight bytes.
Two recording schemes are common one use seven bit bytes per character and other using nine bit bytes. This means that the magnetic tape surface is divided into seven or nine tracks.The area in which data and information are stored on magnetic tape. By Zafar Manzoor
Two billion bytes or 2,000,000,000 bytes is equivalent to:1,953,125 KB~1,907.34 MB~1.86 GB~0.001819 TB
start input A & B if A>B print A is greatest if B>A print B is greatest stop james ola writes.....SOT.
A double byte is two bytes.
in java, char consumes two bytes because it uses unicode instead of ascii. and int takes 4 bytes because 32-bit no will be taken