Answer:
def find_max(num_1, num_2):
max_val = 0.0
if (num_1 > num_2): # if num1 is greater than num2,
max_val = num_1 # then num1 is the maxVal.
else: # Otherwise,
max_val = num_2 # num2 is the maxVal
return max_val
max_sum = 0.0
num_a = float(input())
num_b = float(input())
num_y = float(input())
num_z = float(input())
max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)
print('max_sum is:', max_sum)
Explanation:
I added the missing part. Also, you forgot the put parentheses. I highlighted all.
To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)
A burn-in test is a test which is usually performed on a system or component by running it for a long time in order to bring out any errors or system failures etc.
While doing it on CPU the data must be backed up as any kind of error or failure may result in the loss of data, at time systems can be repaired to retrieve data but still there is no guarantee, backing up is the best option.
Option A because the environment should be selected for which the changes are to be applied. In a time Force IDE has many project environments for example maybe a java project and C++ project would be there on sandbox, so the environment selection is important.
Option B because the related changed sets should be specified so that other developers that have access to the project can see the changes being made.
Option D The data fields that are needed to be deployed should also be provided so that the updated version can be seen by other developers.
Rejected Options :
Option C user name and password has nothing to do with the production environment because if the user has it only then it can come and make changes.
The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.
#include<stdio.h>
#include<string.h>
int* uniqueValue(int input1,int input2[])
{
int left, current;
static int arr[4] = {0};
int i = 0;
for(i=0;i<input1;i++)
{
current = input2[i];
left = 0;
if(current > 0)
left = arr[(current-1)];
if(left == 0 && arr[current] == 0)
{
arr[current] = input1-current;
}
else
{
for(int j=(i+1);j<input1;j++)
{
if(arr[j] == 0)
{
left = arr[(j-1)];
arr[j] = left - 1;
}
}
}
}
return arr;
}
