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java - 给定最大长度求毕达哥拉斯三元和

转载 作者:塔克拉玛干 更新时间:2023-11-02 19:17:08 28 4
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最近我无法确定我遇到的问题的解决方案。我的代码在此过程中跳过了某些毕达哥拉斯三元组,因此给出了不正确的总和。

例如,如果我给最大长度 15(斜边不能大于这个数字)。

它将打印:

3-4-5

5-12-13

然而,存在三个最大边长为 15 的毕达哥拉斯三元组:

3-4-5

8-6-10 - 我错过了这个 :(

5-12-13

我理解手头的问题,但我不确定如何创建有效的解决方案(不创建不必要的循环)。也许有人可以为我指出正确的方向,并在可能的情况下简化我的代码。我总是愿意学习!

class toymeister{

public static void main(String args[]){
int inputLength = 15;

System.out.println("Given the length: " + inputLength +
" the sum of the pythagorean triplets in the given range is: "
+ sumTripletsGivenLength(inputLength));
}

public static int sumTripletsGivenLength(int length){
boolean isFinished = false;
int m,n = 0;
int sum=0;

while(!isFinished){
n++; {
m=n+1; {

int a,b,c;

a = m*m - n*n;
b = 2*m*n;
c = m*m + n*n;

if(c<length){
System.out.println(a + "-" + b + "-" + c);
sum = sum+a+b+c;
} else {
isFinished = true;
}
}
}
}
return sum;
}
}

最佳答案

这是您的 sumTripletsGivenLength() 方法的有效实现:

public static int sumTripletsGivenLength(int length){
int sum = 0;

// the two for loops below are structured such that duplicate pairs
// of values, e.g. (3, 4) and (4, 3), do not occur
for (int a=1; a < length; ++a) {
for (int b=a; b < length; ++b) {
// calculate hypotenuse 'c' for each combintation of 'a' and 'b'
double c = Math.sqrt(Math.pow(a, 2.0) + Math.pow(b, 2.0));

// check if 1) 'c' be a whole number, and 2) its length is within range
if (c == Math.ceil(c) && c < length) {
sum = a + b + (int)c;
System.out.println(a + "-" + b + "-" + (int)c);
}
}
}

return sum;
}

public static void main(String args[]) {
int inputLength = 15;

System.out.println("Given the length: " + inputLength +
" the sum of the pythagorean triplets in the given range is: "
+ sumTripletsGivenLength(inputLength));
}

输出:

3-4-5
5-12-13
6-8-10
Given the length: 15 the sum of the pythagorean triplets in the given range is: 24

关于java - 给定最大长度求毕达哥拉斯三元和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33819049/

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