joj1611 Perfect Cubes

 1611: Perfect Cubes

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	780	320	Standard

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  , 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  (e.g. a quick calculation will show that the equation  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  .

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.

 

 

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This problem is used for contest: 109 

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PS： 强烈鄙视最后的Note...写得貌似这题很容易超时，要仔细从数学上优化一下...结果四个for果断水过了...

#include <stdio.h>

int main()
{
    double cube, a, b, c;

    for (cube=6; cube<=200; ++cube)
    {
        for (a=2; a<cube; ++a)
        {
            for (b=a; b<cube; ++b)
            {
                for (c=b; c<cube; ++c)
                {
                    if (cube*cube*cube == a*a*a + b*b*b + c*c*c)
                    {
                        printf("Cube = %.0lf, Triple = (%.0lf,%.0lf,%.0lf)\n", cube, a, b, c);

                    }
                }
            }
        }
    }

    return 0;
}