Project Euler Problem 75
Project Euler Problem 75
BlackJoker 发表于2年前
Project Euler Problem 75
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It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L  1,500,000 can exactly one integer sided right angle triangle be formed?

Note: This problem has been changed recently, please check that you are using the right parameters.

下面是java解题方法:
public static void main(String[] args) {
		Map<Integer, ABC> dict = new HashMap<Integer, ABC>();
		int L = 1500000;
		int M =  (int)Math.sqrt(L / 2);
		for (int m = 1; m <= M; m++) {
			for (int n = 1; n < m; n++) {
				int a = m * m - n * n;
				int b = 2 * m * n;
				int c = m * m + n * n;
				int l = a + b + c;
				int k = L / l;
				for (int i = 1; i <= k; i++) {
					int[] abc = { i * a, i * b, i * c };
					Arrays.sort(abc);
					check(i, l, abc, dict);
				}
			}
		}
		int count = 0;
		for (ABC abc : dict.values()) {
			count += abc.got;
		}
		System.out.println(count);
	}

	private static void check(int i, int l, int[] abc, Map<Integer, ABC> dict) {
		int ll = i * l;
		ABC tmp = dict.get(ll);
		if (tmp != null) {
			if (tmp.got == 1) {
				int[] tmp1 = tmp.abc;
				for (int x = 0; x < 3; x++) {
					if (tmp1[x] != abc[x]) {
						tmp.got = 0;
						break;
					}
				}
			}
		} else {
			dict.put(ll, new ABC(abc));
		}
	}

	static class ABC {
		public ABC(int[] abc) {
			this.abc = abc;
		}

		int got = 1;
		int[] abc;

	}

这个解法的时间在秒级,解法利用了下面勾股数的性质:
1,通过a = m^2 − n^2, b = 2mn, c = m^2 + n^2 (1)可以生成所有的素勾股数和一部分派生勾股数(a,b,c),其在(m>n,m,n为正整数);
2,勾股数分为素勾股数和派生勾股数,而且所有派生勾股数都是由素勾股数派生而来的;
3,根据1,2,利用式(1),遍历m,n并加上派生规则可以生成所有勾股数。

注:正整数a,b,c满足a^2+b^2=c^2<=>(a,b,c)是勾股数,如果(a,b,c)=1(即a,b,c互素),则(a,b,c)称为素勾股数,对于整数k,(ka,kb,kc)是由(a,b,c)派生的勾股数。
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