Average of n numbers

curious (10)
cin>> n numbers (but there is not exactly n it depends on user how many number he/she will include) n>=2&&n<=1000
how can i find average of them..
Maese909 (131)
To find the sum of n numbers, add all the numbers and divide the sum by n
Last edited on
curious (10)
how ?
i want to write
for (i=2:i<=n;i++)
cin>>a[i]
then sum/n however we cant include n and i cannot do
curious (10)
I know logic first-I have to include numbers then i hve to count maybe with k++ then I have to divide sum of numbers to k--but i cant write it in program esp..first stage how i can include numbers without knowing quantity of them
Last edited on
Janlan (90)
...hope that this code can help you :)))

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#include <iostream>
using namespace std;
#define maxLenght 10000
int main()
{

	int lenght;

	cout<<"Enter the lenegt of the array: ";
	cin>>lenght;

	double myArray[maxLenght];

	
	for(int i = 0; i < lenght; i++)
	{
		cout<<"Enter value with index: "<<i<<endl;
		cin>>myArray[i];
	}

	cout<<"\n";

	double avg;
	double sum = 0;

	for(int i = 0; i < lenght; i++)
	{
		sum += myArray[i];

		avg = sum / lenght;
	}

	cout<<"Avg is: "<<avg<<endl;
	


cin.get(); cin.get();
return 0;
}
ne555 (10692)
how i can include numbers without knowing quantity of them
You need a way to know when the user ended the input. Use that as the condition of the loop.
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