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  3. Project Euler #6: Sum square difference
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Project Euler #6: Sum square difference

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274 Discussions

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  • sir_bryan_chong
     + 0 comments

    Here's my solution:

    record = {1:1}
def sumOfFirstNNumbersSquared(n):
    if n in record:
        return record[n]
    else:
        record[n] = sumOfFirstNNumbersSquared(n - 1) + n ** 2
        return record[n]


t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    
    a = ( n * (n + 1) / 2 ) ** 2
    b = sumOfFirstNNumbersSquared(n)
    print(int( abs(a - b)) )

  • Hannanmuzammil
     + 0 comments
    t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    
    sumofsquare = (n*(n+1)//2)**2 
    squareofsum = n*(n+1)*(2*n+1) // 6 
    print(abs(sumofsquare-squareofsum))

  • surabhimpmattur
     + 0 comments

    why my second case isn't executed? for(int a0 = 0; a0 < t; a0++){ int n; cin >> n; int sum=0; int sqsum=0; int sqaure;

        for(int i=1;i<=n;i++){
         sum+=i;
         sqaure =pow(sum,2);
         sqsum+=pow(i,2);
    }
    int diff=sqaure-sqsum;
     cout<<diff<<endl;
}

  • vedikagupta0
     + 0 comments

    why my second case isn't executed? t = int(input().strip())

    for a0 in range(t):

    n = int(input().strip())

summ_bracket = 0

sum_s = 0

for i in range(1,n+1):

    summ_bracket+=i

    sum_s += i**2

print((summ_bracket)**2-sum_s)

  • tjmfollosco
     + 0 comments

    Had to look up some formulas and do some algebra.

    long ans = n * (n + 1) * (3 * n + 2) * ( n - 1) / 12;