Answer:g
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
Explanation:
Using Java programming language:
- The method addOddMinusEven() is created to accept two parameters of ints start and end
- Using a for loop statement we iterate from start to end but not including end
- Using a modulos operator we check for even and odds
- The method then returns odd-even
- See below a complete method with a call to the method addOddMinusEven()
public class num13 {
public static void main(String[] args) {
int start = 2;
int stop = 10;
System.out.println(addOddMinusEven(start,stop));
}
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
As far as windows 7 and below it was control panel, but in windows 8 and higher, microsoft added, a new app which was called settings, control panel was still there but with a few features removed
Answer:
Kasiski’s method for determining 't' works for Vigenère cipher as well. The only difference is therefore in the second stage of the attack. In the second stage, one needs to build a frequency table for each of the 't' keys, and carry out an attack like on the mono-alphabetic cipher. Given a long enough plaintext, this will work successfully.
Explanation:
Kasiski method is a method of attacking polyalphabetic substitution ciphers such as Vigenère cipher. It is also called Kasiski test or Kasiski examination.
The method involve finding the length of the keyword and then dividing the message into that many simple substitution cryptograms. Frequency analysis could then be used to solve the resulting simple substitution.
8 bits = 1 byte
10 Mbps = 10/8 M bytes per second
1 frame = 50 bytes
Frame rate = 10 M / (8 x 50) = 10 M / 400 = <em>25,000 frames per second</em>
========================
If each frame has an error probability of 0.004, then (0.004 x 25,000) = 100 frames per second are damaged.
1 hour = 3,600 seconds.
100 damaged frames per second = (100 x 3,600) = <em>360,000 damaged frames per hour.</em>
