**I would like to thank all the participants for their participation hope you enjoyed the problems , I'm Really Shockeproud to reach about 100 participants joined the contest**↵
[Problems](https://codeforces.net/contestInvitation/5c0e761de9e4ccb3e94e9986db6567b970647d40)↵
↵
↵
[A. Cubes](https://codeforces.net/gym/469486/problem/A)↵
↵
<spoiler summary="Idea">↵
The Sum $\sum_{x=1}^{n}x^3$ has a closed form which is $(\frac{n(n+1)}{2})^2$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
x=int(input())↵
print(x*(x+1)*x*(x+1)//4)↵
~~~~~↵
</spoiler>↵
↵
[B. Missing Number](https://codeforces.net/gym/469486/problem/A)↵
↵
<spoiler summary="Idea">↵
since the sum of numbers from $1$ to $n$ has a closed form $\frac{n(n+1)}{2}$ we can obtain the missing number by sum all the numbers from ${1}$ to ${n}$ (inclusive) and then sum the given numbers ${x_1,x_2,x_3...,x_{n-1}}$ and find the difference so the final solution is $\frac{n(n+1)}{2}-\sum_{i=1}^{n-1}x_i$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=list(map(int,input().split()))↵
print(n*(n+1)//2-sum(x))↵
~~~~~↵
</spoiler>↵
↵
[C. Max Sub Array](https://codeforces.net/gym/469486/problem/C)↵
↵
<spoiler summary="Idea">↵
The Idea Here is Linearize the Time Complexity by using efficient popular algorithm called "Kadane's Algorithm" that maximize the sum in each iteration to get with the maximum subarray in $O(n)$ time complexity↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=list(map(int,input().split()))↵
sm,best=0,0↵
for i in range(len(x)):↵
sm=max(x[i],sm+x[i])↵
best=max(best,sm)↵
print(best)↵
~~~~~↵
</spoiler>↵
↵
[D. PPV or math](https://codeforces.net/gym/469486/problem/D)↵
↵
<spoiler summary="Idea">↵
The Idea is to (string add) numbers even with themselves , so if we have n numbers then we will have $n*n=n^2$ (string add) operations that means that Time Complexity Here will Be $O(n^2)$ and of course in the iteration converting the string to integer and adding to the sum ↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=input().split()↵
ar=[]↵
for i in range(n):↵
for j in range(n):↵
ar.append(x[i]+x[j])↵
ar=list(map(int,ar))↵
print(sum(ar))↵
~~~~~↵
</spoiler>↵
↵
[E. Teams](https://codeforces.net/gym/469486/problem/E)↵
↵
<spoiler summary="Idea">↵
The Idea here is to find how many (Integer-Pairs) we can make using number $n$ , most of solution used *//* operation which can lies on large values so it's preferred to use *math* library which is already designed well to hold large values so *floor* is better choice so that answer is $\lfloor \frac{x}{2} \rfloor$ ↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
from math import *↵
print(floor(int(input())/2))↵
~~~~~↵
</spoiler>↵
↵
Hope You Guys Enjoyed The Contest Stay Tuned For More !↵
*Best Wishes*
[Problems](https://codeforces.net/contestInvitation/5c0e761de9e4ccb3e94e9986db6567b970647d40)↵
↵
↵
[A. Cubes](https://codeforces.net/gym/469486/problem/A)↵
↵
<spoiler summary="Idea">↵
The Sum $\sum_{x=1}^{n}x^3$ has a closed form which is $(\frac{n(n+1)}{2})^2$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
x=int(input())↵
print(x*(x+1)*x*(x+1)//4)↵
~~~~~↵
</spoiler>↵
↵
[B. Missing Number](https://codeforces.net/gym/469486/problem/A)↵
↵
<spoiler summary="Idea">↵
since the sum of numbers from $1$ to $n$ has a closed form $\frac{n(n+1)}{2}$ we can obtain the missing number by sum all the numbers from ${1}$ to ${n}$ (inclusive) and then sum the given numbers ${x_1,x_2,x_3...,x_{n-1}}$ and find the difference so the final solution is $\frac{n(n+1)}{2}-\sum_{i=1}^{n-1}x_i$↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=list(map(int,input().split()))↵
print(n*(n+1)//2-sum(x))↵
~~~~~↵
</spoiler>↵
↵
[C. Max Sub Array](https://codeforces.net/gym/469486/problem/C)↵
↵
<spoiler summary="Idea">↵
The Idea Here is Linearize the Time Complexity by using efficient popular algorithm called "Kadane's Algorithm" that maximize the sum in each iteration to get with the maximum subarray in $O(n)$ time complexity↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=list(map(int,input().split()))↵
sm,best=0,0↵
for i in range(len(x)):↵
sm=max(x[i],sm+x[i])↵
best=max(best,sm)↵
print(best)↵
~~~~~↵
</spoiler>↵
↵
[D. PPV or math](https://codeforces.net/gym/469486/problem/D)↵
↵
<spoiler summary="Idea">↵
The Idea is to (string add) numbers even with themselves , so if we have n numbers then we will have $n*n=n^2$ (string add) operations that means that Time Complexity Here will Be $O(n^2)$ and of course in the iteration converting the string to integer and adding to the sum ↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
n=int(input())↵
x=input().split()↵
ar=[]↵
for i in range(n):↵
for j in range(n):↵
ar.append(x[i]+x[j])↵
ar=list(map(int,ar))↵
print(sum(ar))↵
~~~~~↵
</spoiler>↵
↵
[E. Teams](https://codeforces.net/gym/469486/problem/E)↵
↵
<spoiler summary="Idea">↵
The Idea here is to find how many (Integer-Pairs) we can make using number $n$ , most of solution used *//* operation which can lies on large values so it's preferred to use *math* library which is already designed well to hold large values so *floor* is better choice so that answer is $\lfloor \frac{x}{2} \rfloor$ ↵
</spoiler>↵
↵
↵
<spoiler summary="Solution">↵
~~~~~↵
from math import *↵
print(floor(int(input())/2))↵
~~~~~↵
</spoiler>↵
↵
Hope You Guys Enjoyed The Contest Stay Tuned For More !↵
*Best Wishes*
