Answer:
Display as
Explanation:
Need the same question, but that's my guess, because display, means to show.
On the one hand it looks pretty cool, on the other hand change is good
Answer:
For this i will use my own PC.
OS - Windows 10
Storage Capacity - 512 GBs
Memory - 16 GB
Wi-Fi - Ethernet
Installed Application - FireFox
Explanation:
An OS is the interface your computer uses.
Storage capacity is the space of your hard drive.
Memory is how much RAM (Random Access Memory) you have
Wi-Fi connectivity is for how your computer connects the the internet.
An installed application is any installed application on your computer.
In this program, I am using the school-based grading system and the program should accept the subject and the number of students.
Program approach:-
- Using the necessary header file.
- Using the standard I/O namespace function.
- Define the main function.
- Declare the variable.
- Display enter obtain marks in 5 subjects.
- Return the value.
Program:-
//header file
#include<iostream>
//using namespace
using namespace std;
//main method
int main()
{
//declare variable
int j;
float mark, sum=0, a;
//display enter obtain marks in 5 subjects
cout<<"Enter Marks obtained in 5 Subjects: ";
for(j=0; j<5; j++)
{
cin>>mark;
sum = sum+mark;
}
a = sum/5;
//display grade
cout<<"\nGrade = ";
if(a>=91 && a<=100)
//display a1
cout<<"a1";
else if(a>=81 && a<91)
//display a2
cout<<"a2";
else if(a>=71 && a<81)
cout<<"b1";
else if(a>=61 && a<71)
cout<<"b2";
else if(a>=51 && a<61)
//display c1
cout<<"c1";
else if(a>=41 && a<51)
//display c2
cout<<"c2";
else if(a>=33 && a<41)
//display d
cout<<"d";
else if(a>=21 && a<33)
//display e1
cout<<"e1";
else if(a>=0 && a<21)
//display e2
cout<<"e2";
else
//display invalid
cout<<"Invalid!";
cout<<endl;
//return the value
return 0;
}
Learn more grading system
brainly.com/question/24298916
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.
