Defined a function that calculates the twice of sum of first N natural numbers as sum(N). Your task is to modify the function to sumX(N, M, K) that calculates sum( K + sum( K + sum( K + …sum(K + N)…))), continuing for M terms. For a given N, M and K calculate the value of sumX(N, M, K).
Note: Since the answer can be very large, print the answer in modulo 10^9 + 7.
Examples:
Input: N = 1, M = 2, K = 3
Output: 552
For M = 2
sum(3 + sum(3 + 1)) = sum(3 + 20) = 552.
Input: N = 3, M =3, K = 2
Output: 1120422
For M = 3
sum(2 + sum(2 + sum(2 + 3))) = sum(2 + sum(2 + 30)) = sum(2 + 1056) = 1120422.
Approach:
- Calculate value of sum(N) using the formula N*(N + 1).
- Run a loop M times, each time adding K to the previous answer and applying sum(prev_ans + K), modulo 10^9 + 7 each time.
- Print the value of sumX(N, M, K) in the end.
Below is the implementation of the above approach:
// C++ program to calculate the // terms of summing of sum series #include <iostream> using namespace std;
# define MOD 1000000007 // Function to calculate // twice of sum of first N natural numbers long sum( long N){
long val = N * (N+1);
val = val % MOD;
return val;
} // Function to calculate the // terms of summing of sum series int sumX( int N, int M, int K){
for ( int i = 0; i < M; i++) {
N = ( int )sum(K + N);
}
N = N % MOD;
return N;
} // Driver Code int main()
{ int N = 1, M = 2, K = 3;
cout << sumX(N, M, K) << endl;
return 0;
} // This code is contributed by Rituraj Jain |
// Java program to calculate the // terms of summing of sum series import java.io.*;
import java.util.*;
import java.lang.*;
class GFG {
static int MOD = 1000000007 ;
// Function to calculate
// twice of sum of first N natural numbers
static long sum( long N)
{
long val = N * (N + 1 );
// taking modulo 10 ^ 9 + 7
val = val % MOD;
return val;
}
// Function to calculate the
// terms of summing of sum series
static int sumX( int N, int M, int K)
{
for ( int i = 0 ; i < M; i++) {
N = ( int )sum(K + N);
}
N = N % MOD;
return N;
}
// Driver code
public static void main(String args[])
{
int N = 1 , M = 2 , K = 3 ;
System.out.println(sumX(N, M, K));
}
} |
# Python3 program to calculate the # terms of summing of sum series MOD = 1000000007
# Function to calculate # twice of sum of first N natural numbers def Sum (N):
val = N * (N + 1 )
# taking modulo 10 ^ 9 + 7
val = val % MOD
return val
# Function to calculate the # terms of summing of sum series def sumX(N, M, K):
for i in range (M):
N = int ( Sum (K + N))
N = N % MOD
return N
if __name__ = = "__main__" :
N, M, K = 1 , 2 , 3
print (sumX(N, M, K))
# This code is contributed by Rituraj Jain |
// C# program to calculate the // terms of summing of sum series using System;
class GFG {
static int MOD = 1000000007;
// Function to calculate
// twice of sum of first N natural numbers
static long sum( long N)
{
long val = N * (N + 1);
// taking modulo 10 ^ 9 + 7
val = val % MOD;
return val;
}
// Function to calculate the
// terms of summing of sum series
static int sumX( int N, int M, int K)
{
for ( int i = 0; i < M; i++) {
N = ( int )sum(K + N);
}
N = N % MOD;
return N;
}
// Driver code
public static void Main()
{
int N = 1, M = 2, K = 3;
Console.WriteLine(sumX(N, M, K));
}
} // This code is contributed by anuj_67.. |
<?php // PHP program to calculate the // terms of summing of sum series // Function to calculate twice of // sum of first N natural numbers function sum( $N )
{ $MOD = 1000000007;
$val = $N * ( $N + 1);
$val = $val % $MOD ;
return $val ;
} // Function to calculate the terms // of summing of sum series function sumX( $N , $M , $K )
{ $MOD = 1000000007;
for ( $i = 0; $i < $M ; $i ++)
{
$N = sum( $K + $N );
}
$N = $N % $MOD ;
return $N ;
} // Driver Code $N = 1;
$M = 2;
$K = 3;
echo (sumX( $N , $M , $K ));
// This code is contributed // by Shivi_Aggarwal ?> |
<script> // Javascript program to calculate the // terms of summing of sum series // Function to calculate twice of // sum of first N natural numbers function sum(N)
{ let MOD = 1000000007;
let val = N * (N + 1);
val = val % MOD;
return val;
} // Function to calculate the terms // of summing of sum series function sumX(N, M, K)
{ let MOD = 1000000007;
for (let i = 0; i < M; i++)
{
N = sum(K + N);
}
N = N % MOD;
return N;
} // Driver Code let N = 1; let M = 2; let K = 3; document.write (sumX(N, M, K)); // This code is contributed // by Sravan </script> |
552
Time Complexity: O(M)
Auxiliary Space: O(1), since no extra space has been taken.