OJ, VJ Perfect Cubes——一道暴搜题。

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 <= 200.
Output
The output should be listed as shown below, one perfect cube per line, in non-decreasing 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 non-decreasing 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.

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)

Note: 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.

Time limit 1000
msMemory
limit32768 k
BOSWindows

What Fuck! 暴搜！

我 TM 还观察了半天样例的规律，找了半天规律写出来的代码还超时了。这叫人情何以堪。以后能暴搜的先试暴搜好了。

看来暴搜也用不了多少时间。

不过，那Time Limit 1000ms是白写的？1s = 1000ms啊。

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
int a, b, c, d;
for(a = 6; a < 201; a++)
for(b = 2; b < a; b++)
for(c = b; c < a - 1; c++)
for(d = c; d < a - 2; d++)
if(a*a*a == b*b*b + c*c*c + d*d*d)
printf("Cube = %d, Triple = (%d,%d,%d)\n", a, b, c, d );
return 0;

}