you have an an integer n such that 1<=n<= 10^12, find the sum of all triagular numbers that are less than or equal n,
time limit: 1 second
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you have an an integer n such that 1<=n<= 10^12, find the sum of all triagular numbers that are less than or equal n,
time limit: 1 second
Название |
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just binary search the index of the last triangular number <=n and use the hockey stick theorem
did not understand hockey stick theorem, could you explain it please? how does it relate to the problem?
sum of first i triangular numbers is the ith tetrahedral number
It can be solved in
O(sqrt(n))
. Triangular numbers isT(k)=k(k+1)/2
, just increment k from 1 whileT(k)<=n
and get a sum ofT(k)
If you have q queries, it can be solved in
O(q*log(n)+sqrt(n))
with precompute prefix sums. Code?And you can use binary search by last
T(k)
,sum of T(k) from 1 to m
ism(m+1)(m+2)/6
And you can solve in
O(1)
by solving equationm * (m+1) / 2 <= n ~ m^2 + m - 2 * n = 0
.Solution:
m = (sqrt(8*n+1)-1)/2
yes please, I would like to see you approach
I like that O(1) query, another thing that could be done here instead of using formula sqrt(8 * n + 1) — 1) / 2; it can be subs. by the following two lines of codes, 1-x=floor(sqrt(2*n)); 2-if(((x+1)*(x))/2>n)--x; x here is the number of the first triangular numbers <= n, then what is left is to calculate the sum of the first x triangular numbers.
x=floor(sqrt(2*n));if(((x+1)*(x))/2>n)--x;
What's going on here? It is attack from one user with 15 accounts?