The decimal form of the 11001101.00111001.10101001.01000010 binary ip address is 205.57.169.66.
An IP address is a binary number with 32 bits. The 32 bits are divided into four octets, which are groupings of 8 bits each. An IP address, however, is shown as a dotted decimal number (for example: 205.57. 32.9).
Memory regions are given binary addresses by the computer system. But in order to access a memory location, the system utilises a certain number of bits. We can address two memory regions with 1 bit. We can address 4 memory locations with 2 bits and 8 memory locations with 3 bits.
The 4 sets of 8 bits in each of the 4 directions are simply written down to create the 32-bit binary IP address.
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Answer:
def occurencesOfWords(words,freq):
frequency_dictionary={}
matched_frequency_words=[]
for i in range(len(words)):
counter=1
a=words[i].lower()
for j in range(len(words)):
if(i==j):
continue
b=words[j].lower()
if(a==b):
counter+=1
frequency_dictionary[words[i]]=counter
for key in frequency_dictionary:
if(frequency_dictionary[key]==freq):
matched_frequency_words.append(key)
return matched_frequency_words
if __name__=='__main__':
user_input=input()
freq=int(input())
words=user_input.split(" ");
required_words=occurencesOfWords(words,freq)
for s in required_words:
print(s)
Explanation:
- Inside the occurencesOfWords function, initialize a frequency_dictionary that is used to store the string and the frequency of that specific string
.
- matched_frequency_words is an array that is used to store each string whose frequency matches with the provided frequency
- Run a nested for loop up to the length of words and inside the nested for loop, skip the iteration if both variables i and j are equal as both strings are at same index.
- Increment the counter variable, if two string are equal.
- Loop through the frequency_dictionary to get the matching frequency strings
.
- Finally test the program inside the main function.
Answer:
public class Main
{
public static void main(String[] args) {
String[] strs = new String[10];
java.util.Scanner sc = new java.util.Scanner(System.in);
for(int i = 0; i < 10; i++){
System.out.print("Enter string " + (i+1) + ":");
strs[i] = sc.nextLine();
}
System.out.println("The strigs with even number of characters is");
for(int i = 0; i < strs.length;i++){
if(strs[i].length() % 2 == 0){
System.out.println(strs[i]);
}
}
}
}
Explanation:
Solution:
The process of transaction can guarantee the reliability of business applications. Locking resources is widely used in distributed transaction management (e.g; two phase commit, 2PC) to keep the system consistent. The locking mechanism, however, potentially results in various deadlocks. In service oriented architecture, the deadlock problem becomes even worse because multiple transactions try to lock shared resources in the unexpectable way due to the more randomicity of transaction requests, which has not been solved by existing research results. In this paper, we investigate how to prevent local deadlocks, caused by the resource competition among multiple sub-transactions of a gl obal transaction, and global deadlocks from the competition among different global transactions. We propose a replication based approach to avoid the local deadlocks, and a timestamp based approach to significantly mitigate the global deadlocks. A general algorithm is designed for both local and global deadlock prevention. The experimental results demonstrate the effectiveness and efficiency of our deadlock prevention approach. Further, it is also proved that our approach provides higher system performance than traditional resource allocation schemes.
This is the required answer.
