Hello coders, today we are going to solve Puppy and Sum CodeChef Solution whose Problem Code is PPSUM.

Task
Yesterday, puppy Tuzik learned a magically efficient method to find the sum of the integers from 1 to N. He denotes it as sum(N). But today, as a true explorer, he defined his own new function: sum(D, N), which means the operation sum applied D times: the first time to N, and each subsequent time to the result of the previous operation.
For example, if D = 2 and N = 3, then sum(2, 3) equals to sum(sum(3)) = sum(1 + 2 + 3) = sum(6) = 21.
Tuzik wants to calculate some values of the sum(D, N) function. Will you help him with that?
Input Format
The first line contains a single integer T, the number of test cases. Each test case is described by a single line containing two integers D and N.
Output Format
For each testcase, output one integer on a separate line.
Constraints
- 1 ≤ T ≤ 16
- 1 ≤ D, N ≤ 4
Example
Sample Input
2
1 4
2 3
Sample Output
10
21
Explanation
The first test case: sum(1, 4) = sum(4) = 1 + 2 + 3 + 4 = 10.
The second test case: sum(2, 3) = sum(sum(3)) = sum(1 + 2 + 3) = sum(6) = 1 + 2 + 3 + 4 + 5 + 6 = 21.
Solution – Puppy and Sum | CodeChef Solution
C++
#include <iostream> using namespace::std; int main(int argc, const char * argv[]) { int t; cin >> t; while (t--) { int D,N; cin >> D >> N; for (int j = 1; j <= D; j++) { int ans = 0; for (int i = 1 ; i <= N; i++) { ans += i; } N = ans; } cout << N << endl; } return 0; }
Python
#Solution Provided by Sloth Coders T = int(input()) for _ in range(T): a, b = map(int, input().split()) for i in range(0, a): b = (b *(b +1)) // 2 print(b)
Java
/* package codechef; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Codechef { public static void main (String[] args) throws java.lang.Exception { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); int s = 0; for(int i=0; i<t; i++) { int n1 = sc.nextInt(); int n2 = sc.nextInt(); for(int x=1; x<=n1; x++) { s = 0; for(int j=1; j<=n2; j++) { s += j; } n2 = s; } System.out.println(s); } sc.close(); } }
Disclaimer: The above Problem (Puppy and Sum) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.