ZOJ Problem Set

Perfect Cubes

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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. 

解题报告：

一直以为输入一个数求解d^3=a^3+b^3+c^3,发现一直WR，最后看了别人的思路，才知道直接枚举200以下所有可能，真是醉了！

#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;}