我目前正在尝试这个问题:

A Pythagorean triplet is a set of three natural numbers, a, b and c, for which a² + b² = c².

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

我的代码如下，我认为它应该是正确的，但是网站告诉我我的答案是错误的？有人可以帮我看看我逻辑上的缺陷吗？

public class Pythagoras {
    public static void main(String[] args) {
            int sum = 1000;
            int a;
            int product=0;
            for (a = 1; a <= sum/3; a++)
            {
                int b;
                for (b = a + 1; b <= sum/2; b++)
                {
                    int c = sum - a - b;
                    if ( c > 0 && (a*a + b*b == c*c) )
                       System.out.printf("a=%d, b=%d, c=%d\n",a,b,c);
                        product = a * b * c;
                }
            }
            System.out.println(product);
        }
    }

最佳答案

这里有 5 个解决方案(从慢到快):

1) 简单的实现 - 732857 微秒(0.7 秒)

private static void p1(int sum) {
    for (int a = 0; a <= sum; a++) {
        for (int b = 0; b <= sum; b++) {
            for (int c = 0; c <= sum; c++) {
                if (a < b && b < c && a + b + c == sum
                        && (c * c == a * a + b * b)) {
                    System.out.print(a * b * c);
                    return;
                }
            }
        }
    }
}

2)限制b&c的下界(建立顺序关系)-251091微秒(0.2秒)

private static void p2(int sum) {
    for (int a = 0; a <= sum; a++) {
        for (int b = a + 1; b <= sum; b++) {
            for (int c = b + 1; c <= sum; c++) {
                if (a + b + c == sum && (c * c == a * a + b * b)) {
                    System.out.print(a * b * c);
                    return;
                }
            }
        }
    }
}

3) 限制 b 和 c 的下限和上限 - 111220 微秒(0.1 秒)

private static void p3(int sum) {
    for (int a = 0; a <= sum; a++) {
        for (int b = a + 1; b <= sum - a; b++) {
            for (int c = b + 1; c <= sum - a - b; c++) {
                if (a + b + c == sum && (c * c == a * a + b * b)) {
                    System.out.print(a * b * c);
                    return;
                }
            }
        }
    }
}

4) 限制 b 的下限和上限并固定 c 的值 - 2625 微秒

private static void p4(int sum) {
    for (int a = 0; a <= sum; a++) {
        for (int b = a + 1; b <= sum - a; b++) {
            int c = sum - a - b;
            if (c > b && c * c == a * a + b * b) {
                System.out.print(a * b * c);
                return;
            }
        }
    }
}

5) 使用Euclid 公式 - 213 微秒

private static void p5(int sum) {
    // a = m^2 - n^2
    // b = 2mn
    // c = m^2 + n^2
    int a, b, c;
    int sqrt = (int)Math.sqrt(sum);
    for (int n = 1; n <= sqrt; n++) {
        for (int m = n+1; m <= sqrt; m++) {
            a = m*m - n*n;
            b = 2*m*n;
            c = m*m + n*n;
            if ( a + b + c == 1000 ) {
                System.out.print(a * b * c);
                return;
            }
        }
    }
}

关于java - 寻找毕达哥拉斯三胞胎的代码，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7704673/