0xffffffffffffffff
As an unsigned 64-bit integer, this represents the value 18,446,744,073,709,551,615.
However, as a signed 64-bit integer, this only represents the value -1. The signed range is -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 making 0x7fffffffffffffff the largest possible positive value, and 0x8000000000000000 the smallest possible negative value.
Assume the first element is the largest and store its value. Then traverse the remainder of the list, one element at a time. If the current element's value is larger than the stored value, overwrite the stored value with the current element's value. Once you've traversed the list, the stored value will hold the largest value.
12.
Hexadecimal color values are representative of variations in Red, Green and Blue (RGB) color, within the structure of the Hexadecimal, or Base 16 positional numeral system. Within this system, values are represented 0-9, and A-F (representing 11-16). The structure of hexadecimal color value is actually three sets of two digit hexadecimal numbers (e.g. 0F 28 7D). The first value set determines the amount of red, the second green, and the third blue. Within each set, the first number is in value increments of 16, and the second in increments of 1. Where "12 34 56" represents 18 levels of red, 52 levels of green, and 36 levels of blue. You can get a pretty good feel for this color system by playing with the tool over at ColourLovers (see related link.)
import java.util.Scanner; public class NumberSystem { public void displayConversion() { Scanner input = new Scanner(System.in); System.out.printf("%-20s%-20s%-20s%-20s\n", "Decimal", "Binary", "Octal", "Hexadecimal"); for ( int i = 1; i <= 256; i++ ) { String binary = Integer.toBinaryString(i); String octal = Integer.toOctalString(i); String hexadecimal = Integer.toHexString(i); System.out.format("%-20d%-20s%-20s%-20s\n", i, binary, octal, hexadecimal); } } // returns a string representation of the decimal number in binary public String toBinaryString( int dec ) { String binary = " "; while (dec >= 1 ) { int value = dec % 2; binary = value + binary; dec /= 2; } return binary; } //returns a string representation of the number in octal public String toOctalString( int dec ) { String octal = " "; while ( dec >= 1 ) { int value = dec % 8; octal = value + octal; dec /= 8; } return octal; } public String toHexString( int dec ) { String hexadecimal = " "; while ( dec >= 1 ) { int value = dec % 16; switch (value) { case 10: hexadecimal = "A" + hexadecimal; break; case 11: hexadecimal = "B" + hexadecimal; break; case 12: hexadecimal = "C" + hexadecimal; break; case 13: hexadecimal = "D" + hexadecimal; break; case 14: hexadecimal = "E" + hexadecimal; break; case 15: hexadecimal = "F" + hexadecimal; break; default: hexadecimal = value + hexadecimal; break; } dec /= 16; } return hexadecimal; } public static void main( String args[]) { NumberSystem apps = new NumberSystem(); apps.displayConversion(); } }
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
Largest 8 bit unsigned number is 11111111 binary which is the number 255 in decimal. In hexadecimal 255 is represented as FF In octal 255 is represented as 377. The related link below will help.
15
255
Assume the first element is the largest and store its value. Then traverse the remainder of the list, one element at a time. If the current element's value is larger than the stored value, overwrite the stored value with the current element's value. Once you've traversed the list, the stored value will hold the largest value.
409610
162 - 1 = 255 Strictly speaking the highest value is FF which, in decimal is 1515 = 4.38*1017 or approx 438 quadrillion.
The value in hexadecimal of the decimal number 999910 is F41E6.
DCE means: Data Carrier Equipment Hexadecimal value in decimal: 3192
999,999
12.
It is -F.
It is FFFFFF.