UVa 264

264 - Count on Cantor

• http://acm.uva.es/p/v2/264.html

Summary

No point in brute forcing this thing 10^7 will take a long time :)

Explanation

• We can generalize this pattern in order to figure out the fraction.
• Try to sum up all the numbers sequentially until you have figured out the range in which your number will belong.
• Once we establish the range, you can simply figure out the distance from the current position and finish off the problem.
• Follow this general algorithm. Pre-compute an array of summations 4500 in length. Binary search to find exactly what diagonal in the matrix would be referenced. Next, note the when to add/subtract from the numerator and denominators. Easy solution from there.

fzort: This problem also has a O(1) solution. It's easy to compute the index d of the diagonal a given number n is in. The number of elements before diagonal d is

$n=\frac{d(d + 1)}{2}$

, so the number n is in diagonal

$d=\left \lfloor \frac{1 + \sqrt{1+8n}}{2} \right \rfloor$

Input

1
3
14
7
10000000

Output

TERM 1 IS 1/1
TERM 3 IS 2/1
TERM 14 IS 2/4
TERM 7 IS 1/4
TERM 10000000 IS 2844/1629