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Answer:
By Using the Greedy- Activity- Selection algorithm
Explanation:
The Greedy- Activity- Selection algorithm in this case involves
First finding a maximum size set S1, of compatible activities from S for the first lecture hall.
Then using it again to find a maximum size set S2 of compatible activities from S - S1 for the second hall.
This is repeated till all the activities are assigned.
It requires θ(n2) time in its worse .
Answer:
// program in C++.
#include <bits/stdc++.h>
using namespace std;
int main() {
// variable
int num;
cout<<"Enter the number between 20 and 98: ";
// read number
cin >> num;
while(num<20||num>98)
{
cout<<"Wrong input!!enter number between 20-98 only:";
cin>>num;
}
cout<<"The output is: ";
while(num % 10 != num /10)
{
// print numbers.
cout<<num<<" ";
// update num.
num--;
}
// display the number.
cout<<num<<endl;;
return 0;
}
Explanation:
Read a number from user and assign it to variable "num".Check if entered number is in between 20-98 or not.If input number is less than 20 or greater than 98 then ask again to enter a number between 20-98 until user enter a valid input.Then print the countdown from input number till both the digit of number are same.
Output:
Enter the number between 20 and 98: 99
Wrong input!!enter number between 20-98 only:12
Wrong input!!enter number between 20-98 only:93
The output is: 93 92 91 90 89 88
Enter the number between 20 and 98: 77
The output is: 77
Answer:
d the overall strength of colours
Explanation:
Answer:
I wonder what are you saying?
Explanation:
Please give me brainliest :)
