Problem Description

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.

题意：n在[2,200]的范围，都是整数

找出所有的n*n*n=a*a*a+b*b*b+c*c*c;

（<2a<=b<=c<200）

直接暴力做！

注意的只有格式：=号两边都有空格，第一个逗号后面有一个空格。

public class Main{

    public static void main(String[] args) {

        for(int m=6;m<=200;m++){

            int mt = m*m*m;

            int at;

            int bt;

            int ct;

            for(int a=2;a<m;a++){

                at=a*a*a;

                for(int b=a;b<m;b++){

                    bt = b*b*b;

                    //适当的防范一下，提高效率

                    if(at+bt>mt){

                        break;

                    }

                    for(int c=b;c<m;c++){

                        ct=c*c*c;

                        //适当的防范一下，提高效率

                        if(at+bt+ct>mt){

                            break;

                        }

                        if(mt==(at+bt+ct)){

                            System.out.println("Cube = "+m+", Triple = ("+a+","+b+","+c+")");

                        }

                    }

                }

            }

        }

    }

}