Incomplete question. Here's a similar question found in the attachments.
<u>Answer:</u>
<u>General Motors Corp, Chrysler, Ford Motor Company, Toyota Motors sales USA Inc, Nissan North America Inc.</u>
<u>Explanation</u>:
By, using the MS Excel computer software, you would be able to compare the 2016 year-to-date sales through February 2017 year-to-date sales for each manufacturer.
Simply copy the data into the data cells of MS Excel, next use the =sum formula (which should have a minus sign; For example, =SUM (B5 - E5) would give you the difference between the 2016 sales and 2017 sales, only if sales for 2016 is found in column B row 5 and sales for 2017 in column E row 5).
Thus, the results obtained can further be evaluated to determine the manufacturers in the top five with increased sales.
12:21 is the correct answer
Answer: a replay attack, a replay attack is used so that the attacker can go sniff out the hash, and get whatever they are trying to get, then once it goes to the attacker it will go back to the original connection after replaying the hash
Answer:
umm... you did not discribe the anwsers, i cannot help you
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
