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++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the required sumint sum(int A, int B, int R){ // To store the sum int sum = 0; // For every row for (int i = 1; i <= R; i++) { // Update the sum as A appears i number // of times in the current row sum = sum + (i * A); // Update A for the next row A = A + B; } // Return the sum return sum;}// Driver codeint main(){ int A = 5, B = 3, R = 3; cout << sum(A, B, R); return 0;} |
Java
// JAVA implementation of the approachimport java.util.*;import java.lang.*;import java.io.*;class GFG{// Function to return the required sumstatic int sum(int A, int B, int R){ // To store the sum int sum = 0; // For every row for (int i = 1; i <= R; i++) { // Update the sum as A appears i number // of times in the current row sum = sum + (i * A); // Update A for the next row A = A + B; } // Return the sum return sum;}// Driver codepublic static void main (String[] args) throws java.lang.Exception{ int A = 5, B = 3, R = 3; System.out.print(sum(A, B, R));}}// This code is contributed by nidhiva |
Python3
# Python3 implementation of the approach# Function to return the required sumdef Sum(A, B, R): # To store the sum ssum = 0 # For every row for i in range(1, R + 1): # Update the sum as A appears i number # of times in the current row sum = sum + (i * A) # Update A for the next row A = A + B # Return the sum return sum# Driver codeA, B, R = 5, 3, 3print(Sum(A, B, R))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the required sumstatic int sum(int A, int B, int R){ // To store the sum int sum = 0; // For every row for (int i = 1; i <= R; i++) { // Update the sum as A appears i number // of times in the current row sum = sum + (i * A); // Update A for the next row A = A + B; } // Return the sum return sum;}// Driver codepublic static void Main () { int A = 5, B = 3, R = 3; Console.Write(sum(A, B, R));}}// This code is contributed by anuj_67.. |
Javascript
<script>// JAVA SCRIPT implementation of the approach// Function to return the required sumfunction sum( A, B, R){ // To store the sum let sum = 0; // For every row for (let i = 1; i <= R; i++) { // Update the sum as A appears i number // of times in the current row sum = sum + (i * A); // Update A for the next row A = A + B; } // Return the sum return sum;}// Driver code let A = 5, B = 3, R = 3; document.write(sum(A, B, R));//contributed by bobby</script> |
Output:
54
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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
