386 - Perfect Cubes

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# 題目連結

• 題目連結
• Online Judge
• uDebug

# 題目說明

Time limit: 3.000 seconds

# 題目

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)

# 解題技巧

要避免掉 TLE，可以嘗試以 3 個 for 迴圈解題， cubeD = cubeA - cubeB - cubeC 來求出原本要跑第 4 次 for 迴圈的 cubeD ，來減少 time complexity。

# Solution

public class Main {

    public static void main(String[] args) {

        for (int a = 2; a <= 200; a++) {

            int cubeA = (int) Math.pow(a, 3);

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

                int cubeB = (int) Math.pow(b, 3);

                for (int c = b + 1; c < a; c++) {

                    int cubeC = (int) Math.pow(c, 3);

                    int cubeD = cubeA - cubeB - cubeC;

                    if(cubeC > cubeD){

                        continue;

                    }

                    int d = (int)Math.round(Math.pow(cubeD, 1.0/3));

                    if ((int)Math.pow(d, 3) == cubeD) {

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

                    }

                }

            }

        }

    }

}