CUBES  Perfect Cubes
For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the "perfect cube" equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= 100.
The output should be listed as shown below, one perfect cube per line, in nondecreasing order of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in nondecreasing order on the line itself. There do exist several values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first.
Note that the programmer will need to be concerned with an efficient implementation. The official time limit for this problem is 2 minutes, and it is indeed possible to write a solution to this problem which executes in under 2 minutes on a 33 MHz 80386 machine. Due to the distributed nature of the contest in this region, judges have been instructed to make the official time limit at their site the greater of 2 minutes or twice the time taken by the judge's solution on the machine being used to judge this problem.
The first part of the output is shown here:
Cube = 6, Triple = (3,4,5) Cube = 12, Triple = (6,8,10) Cube = 18, Triple = (2,12,16) Cube = 18, Triple = (9,12,15) Cube = 19, Triple = (3,10,18) Cube = 20, Triple = (7,14,17) Cube = 24, Triple = (12,16,20)
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dev_harsh1998:
20190505 12:44:40
lol bruteforce Last edit: 20190505 12:45:03 

kass_97:
20171215 18:59:28
Back to SPOJ after a long time, solved this one first 

nadstratosfer:
20170807 10:48:48
Optimized brute force (about 4M iterations) is enough with Python. 

lt:
20170425 20:41:51
Solved in O(N^2 logN), but no doubt it can be easily solved in O(N^2) 

cake_is_a_lie:
20170306 16:15:56
You can always brute force and submit as TEXT, or even send in the brute force solution. But it's much more elegant to try and solve it in O(N^2 log N). 

suraj_:
20161101 06:46:25
AC!! in one go..use brute force + set; 

vineetpratik:
20160627 09:55:30
100th :) 

mkfeuhrer:
20160604 13:54:40
print all possible solutions just in sorted order ..... 1 WA ...brute force AC:) 

akshayvenkat:
20160307 18:37:19
Brute Force accepted.. 9 = (1,6,8) case fetched me the wrong answer. make sure a,b,c are greater than one, and only then print! Easy stuff! 

dwij28:
20160113 02:58:27
Python results in TLE but got AC with C, C++ and TEXT.. Would love to hear from someone who has an AC python solution.. Did you guys use binary search or something of that sort ? 
Added by:  Wanderley GuimarÄƒes 
Date:  20060601 
Time limit:  0.990s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ERL JSRHINO NODEJS PERL6 VB.NET 
Resource:  ACM Mid Central Regionals 1995 