# Puppy and Sum | CodeChef Solution

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

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.