最近,我一直无法确定解决问题的方法。我的代码在此过程中跳过某些毕达哥拉斯三胞胎,因此给出了不正确的总和。
例如,如果我给出最大长度15(斜边不能大于这个数字)。
它将打印:
3-4-5
13年5月12日
然而,有三个毕达哥拉斯三元组,其给定的最大边长为15:
3-4-5
8-6-10 - 我错过了这个:(
13年5月12日
我理解手头的问题,但我不确定如何创建有效的解决方案(不会产生不必要的循环)。也许有人可以指出我正确的方向,如果可能的话可能会简化我的代码。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;
}
}
答案 0 :(得分:2)
以下是sumTripletsGivenLength()
方法的有效实施:
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));
}
<强>输出:强>
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