21, ఆగస్టు 2025, గురువారం
17, ఆగస్టు 2025, ఆదివారం
8, ఆగస్టు 2025, శుక్రవారం
cyber security.............
Volume-based attacks, also known as volumetric attacks, are a type of DDoS attack that flood a target with massive amounts of network traffic, overwhelming its resources and causing service disruptions. These attacks are typically measured in bits per second (bps), packets per second (pps), or connections per second (cps). The goal is to saturate the target's bandwidth and processing power, making it unable to handle legitimate traffic
Application-based attacks exploit vulnerabilities in software applications to gain unauthorized access, disrupt functionality, or steal data. These attacks can target various layers of an application, including the web, API, and mobile application layers. Common types include injection attacks, cross-site scripting (XSS), cross-site request forgery (CSRF), and denial-of-service (DoS/DDoS) attacks.
A URL interpretation attack, also known as a semantic URL attack, occurs when an attacker manipulates the parameters of a URL to exploit vulnerabilities in how a web application interprets the URL's syntax and semantics. This can lead to unauthorized access to resources, modification of data, or other unexpected behavior
A dictionary attack is a type of cyberattack where hackers try to guess passwords by systematically trying words from a pre-defined list (a "dictionary") of common passwords. This method exploits the fact that many people use predictable or easily guessed passwords. Attackers use automated tools to rapidly input these potential passwords until they find a match, gaining unauthorized access to systems or data
Protocol-based attacks are a type of DDoS (Distributed Denial of Service) attack that exploits weaknesses in network protocols, particularly at Layers 3 and 4 of the OSI model. These attacks aim to disrupt a service by overwhelming a target's resources with malicious connection requests or by exploiting protocol vulnerabilities to exhaust server capacity.
lab-4 python ..................
rows = int(input("Enter number of rows: "))
for i in range(rows):
for j in range(i+1):
print(j+1, end=" ")
print()
// C++ implementation of the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to return the word
// of the corresponding digit
void printValue(char digit)
{
// Switch block to check for each digit c
switch (digit) {
// For digit 0
case '0':
cout << "Zero ";
break;
// For digit 1
case '1':
cout << "One ";
break;
// For digit 2
case '2':
cout << "Two ";
break;
// For digit 3
case '3':
cout << "Three ";
break;
// For digit 4
case '4':
cout << "Four ";
break;
// For digit 5
case '5':
cout << "Five ";
break;
// For digit 6
case '6':
cout << "Six ";
break;
// For digit 7
case '7':
cout << "Seven ";
break;
// For digit 8
case '8':
cout << "Eight ";
break;
// For digit 9
case '9':
cout << "Nine ";
break;
}
}
// Function to iterate through every
// digit in the given number
void printWord(string N)
{
int i, length = N.length();
// Finding each digit of the number
for (i = 0; i < length; i++) {
// Print the digit in words
printValue(N[i]);
}
}
// Driver code
int main()
{
string N = "123";
printWord(N);
return 0;
}
Input: N = 1234
Output: One Two Three Four
Explanation:
Every digit of the given number has been converted into its corresponding word.
#
Python program to determine whether # the number is Armstrong number or not #
Function to calculate x raised to # the power y def power(x, y): if y == 0:
return 1 if y % 2 == 0: return power(x, y // 2) * power(x, y // 2) return x *
power(x, y // 2) * power(x, y // 2) # Function to calculate order of the number
def order(x): # Variable to store of the number n = 0 while (x != 0): n = n + 1
x = x // 10 return n # Function to check whether the given # number is
Armstrong number or not def isArmstrong(x): n = order(x) temp = x sum1 = 0
while (temp != 0): r = temp % 10 sum1 = sum1 + power(r, n) temp = temp // 10 #
If condition satisfies return (sum1 == x) # Driver code x = 153
print(isArmstrong(x)) x = 1253 print(isArmstrong(x))
# Multiplication
table (from 1 to 10) in Python
num = 12
# To take input
from the user
# num =
int(input("Display multiplication table of? "))
# Iterate 10
times from i = 1 to 10
for i in range(1,
11):
print(num, 'x', i, '=', num*i)
Input: N = 567
Output: Five Six Seven
7, ఆగస్టు 2025, గురువారం
6, ఆగస్టు 2025, బుధవారం
ONE DOCTORATE SAID TO RAMU HE USES CHATGPT FOR EVERY TOPIC ...............RAMU ASKED HIM HOW MUCH U PAID TO BUY YOUR DOCTORATE .......my brother with degree is getting 24 lakhs salary per Anam In my department my colleague doctorate is getting 8 lakhs only Anam Big laugh by ramu………….. musth joke ah ah creaming............................................................
One doctorate said to ramu "he uses ChatGPT for every topic"
Ramu asked him how much u paid to buy your doctorate
What a wonderful innocent doctorate I meat I laughed loudly
The remain discussion read between the line
E=mc2 only for student, Scientist or phd.......... ramu quoted this stmt
and for u any change in the above eq.........................
my brother with degree is getting 24 lakhs salary per Anam
In my department my colleague doctorate is getting 8 lakhs
only
Big laugh by ramu…………..
musth joke ah ah Screaming.......................