project euler 75

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Problem 75

Singular integer right triangles

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?

唯一的整数边直角三角形

只能唯一地弯折成整数边直角三角形的电线最短长度是12厘米；当然，还有很多长度的电线都只能唯一地弯折成整数边直角三角形，例如：

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

相反地，有些长度的电线，比如20厘米，不可能弯折成任何整数边直角三角形，而另一些长度则有多个解；例如，120厘米的电线可以弯折成三个不同的整数边直角三角形。

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

记电线长度为L，对于L ≤ 1,500,000，有多少种取值只能唯一地弯折成整数边直角三角形？

package projecteuler;import java.util.HashMap;import java.util.Map;import java.util.Map.Entry;import junit.framework.TestCase;public class Prj75 extends TestCase{private static final int UP_LIMIT = 1500000;/** * a = k * ( 2mn); * b = k * ( mm + nn); * c = k * ( mm - nn); * b + c < p ===> m < sqrt( p/ 2); *  * (m,n) == 1; *  */public void testSingularIntegerRightTriangles(){Map<Integer, Integer> map = new HashMap<Integer, Integer>();int p = UP_LIMIT;for( int m = 2; m <= Math.sqrt(p/2); m ++){for(int n = 1; n < m; n ++){if( gcd(m, n) == 1 && ( m- n) % 2 != 0){for( int k = 2 * m * ( m + n); k < p; k = k + 2 * m * (m + n)){if( !map.containsKey(k)){map.put(k, 0);}map.put(k, map.get(k) + 1);}}}}int count = 0;for( Entry<Integer, Integer> entry : map.entrySet()){if( entry.getValue() == 1){count ++;}}System.out.println("count=" + count);}int gcd(int m, int n) {int a = m;int b = n;if (m < n) {a = n;b = m;}int t = 0;while (a % b != 0) {t = a % b;a = b;b = t;}return b;}}

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