P.E.6: Sum Square Difference

C++ solution:

#include<iostream>

using namespace std;

int main (){
    int S1=0, S2=0;
    int i=1, j=1;

    while(i<=100){
        S1=S1+i*i;
        i++;
    }

    while(j<=100){
        S2=S2+j;
         j++;   
    }

    cout << S2*S2-S1;

    return 0;
}

Java solution (Try online https://ideone.com/onQ94s) -

int sum1 = 0, sum2 = 0, j;
for(int i = 1; i <= 100; i++)
    sum1 += i*i;
for(j = 1; j <= 100; j++)
    sum2 += j;
sum2 *= sum2;

System.out.println(sum2 - sum1);

Naive direct translation in Python:

(sum(range(1,101))**2) - sum([x**2 for x in range(1,101)])

It's simple in Haskell:

1

(sum [1..100])^2 - sum [ x*x | x <- [1..100]] 

My python code is as follows:

# -*- coding: utf-8 -*-
 """
Euler Problem #6:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 55^2 = 3025
 Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
 Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
"""
 prod=0 
 for i in range(1,101):
    square_of_each= i*i
    prod = prod + square_of_each #Sum of squares of first 100 natural numbers
print "The sum of the squares of the first 100 natural numbers is: ", prod
n=100
sum = n*(n + 1)//2
square_of_all= sum*sum
print "The sum of the squares of the first 100 natural numbers is: ", square_of_all
print "The difference between them is: ",square_of_all - prod
raw_input()

Hope this helps!!