Find the sum of elements of the Matrix generated by the given rules
Given three integers A, B, and R, the task is to find the sum of all the elements of the matrix generated by the given rules:
- The first row will contain a single element which is A and the rest of the elements will be 0.
- The next row will contain two elements all of which are (A + B) and the rest are 0s.
- Third row will contain (A + B + B) three times and the rest are 0s and so on.
- The matrix will contain only R rows.
For example, if A = 5, B = 3 and R = 3 then the matrix will be:
5 0 0
8 8 0
11 11 11
Examples:
Input: A = 5, B = 3, R = 3
Output: 54
5 + 8 + 8 + 11 + 11 + 11 = 54
Input: A = 7, B = 56, R = 1
Output: 7
Approach: Initialize sum = 0 and for every 1 ? i ? R update sum = sum + (i * A). After every iteration update A = A + B. Print the final sum in the end.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int sum( int A, int B, int R)
{
int sum = 0;
for ( int i = 1; i <= R; i++) {
sum = sum + (i * A);
A = A + B;
}
return sum;
}
int main()
{
int A = 5, B = 3, R = 3;
cout << sum(A, B, R);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
static int sum( int A, int B, int R)
{
int sum = 0 ;
for ( int i = 1 ; i <= R; i++)
{
sum = sum + (i * A);
A = A + B;
}
return sum;
}
public static void main (String[] args)
throws java.lang.Exception
{
int A = 5 , B = 3 , R = 3 ;
System.out.print(sum(A, B, R));
}
}
|
Python3
def Sum (A, B, R):
ssum = 0
for i in range ( 1 , R + 1 ):
sum = sum + (i * A)
A = A + B
return sum
A, B, R = 5 , 3 , 3
print ( Sum (A, B, R))
|
C#
using System;
class GFG
{
static int sum( int A, int B, int R)
{
int sum = 0;
for ( int i = 1; i <= R; i++)
{
sum = sum + (i * A);
A = A + B;
}
return sum;
}
public static void Main ()
{
int A = 5, B = 3, R = 3;
Console.Write(sum(A, B, R));
}
}
|
Javascript
<script>
function sum( A, B, R)
{
let sum = 0;
for (let i = 1; i <= R; i++)
{
sum = sum + (i * A);
A = A + B;
}
return sum;
}
let A = 5, B = 3, R = 3;
document.write(sum(A, B, R));
</script>
|
Time Complexity: O(R), since there runs a loop for once from 1 to R.
Auxiliary Space: O(1), since no extra space has been taken.
Last Updated :
28 May, 2022
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