【Codewars】Integers: Recreation One
Codewars里的 5kyu Kata。
题目说明：
Divisors of 42 are : 1, 2, 3, 6, 7, 14, 21, 42. These divisors squared are: 1, 4, 9, 36, 49, 196, 441, 1764. The sum of the squared divisors is 2500 which is 50 * 50, a square!
Given two integers m, n (1 <= m <= n) we want to find all integers between m and n whose sum of squared divisors is itself a square. 42 is such a number.
The result will be an array of arrays or of tuples (in C an array of Pair) or a string, each subarray having two elements, first the number whose squared divisors is a square and then the sum of the squared divisors.
#Examples:
list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]] list_squared(42, 250) --> [[42, 2500], [246, 84100]]The form of the examples may change according to the language, see
Example Tests:for more details.Note
In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.
解题代码：
import java.util.ArrayList;
public class SumSquaredDivisors {
public static String listSquared(long m, long n) {
// your code
ArrayList divisors = new ArrayList();
ArrayList> res = new ArrayList>();
for (long c = m; c < n; c++) {
divisors.clear();
//divisors.add((long) 1);
for (long i = 1; i <= c / 2; i++) {
if (c % i == 0) {
divisors.add(i);
}
}
divisors.add(c);
long sum = new Long(0);
for (int i = 0; i < divisors.size(); i++) {
sum += (long) Math.pow(divisors.get(i).longValue(), 2);
}
Long s = new Long(sum);
Double d = new Double(Math.sqrt(sum));
if (d.toString().substring(d.toString().indexOf('.')).equals(".0")) {
ArrayList r = new ArrayList();
r.add(c);
r.add(sum);
res.add(r);
}
}
return res.toString();
}
} Test Cases:
import static org.junit.Assert.*;
import java.util.Random;
import org.junit.Test;
public class SumSquaredDivisorsTest {
private static long[] solAux(long n) {
long s = 0;
long nf = 0;
long[] res = new long[2];
for (long i = 1; i <= Math.floor(Math.sqrt(n)); i += 1) {
if (n % i == 0) {
s += i * i;
nf = n / i;
if (nf != i)
s += nf * nf;
}
}
if (Math.pow((long) Math.sqrt(s), 2) == s) {
res[0] = n;
res[1] = s;
return res;
} else
return null;
}
public static String solution(long m, long n) {
String res = "[";
for (long i = m; i <= n; i++) {
long[] r = solAux(i);
if (r != null) {
res += "[" + Long.toString(r[0]) + ", " + Long.toString(r[1]) + "], ";
}
}
return res.replaceFirst(",\\s$", "") + "]";
}
@Test
public void test1() {
assertEquals("[[1, 1], [42, 2500], [246, 84100]]", SumSquaredDivisors.listSquared(1, 250));
}
@Test
public void test2() {
assertEquals("[[42, 2500], [246, 84100]]", SumSquaredDivisors.listSquared(42, 250));
}
@Test
public void test3() {
assertEquals("[[287, 84100]]", SumSquaredDivisors.listSquared(250, 500));
}
@Test
public void test4() {
assertEquals("[]", SumSquaredDivisors.listSquared(300, 600));
}
@Test
public void test5() {
assertEquals("[[728, 722500], [1434, 2856100]]", SumSquaredDivisors.listSquared(600, 1500));
}
@Test
public void test6() {
assertEquals("[[1673, 2856100]]", SumSquaredDivisors.listSquared(1500, 1800));
}
@Test
public void test7() {
assertEquals("[[1880, 4884100]]", SumSquaredDivisors.listSquared(1800, 2000));
}
@Test
public void test8() {
assertEquals("[]", SumSquaredDivisors.listSquared(2000, 2200));
}
@Test
public void test9() {
assertEquals("[[4264, 24304900]]", SumSquaredDivisors.listSquared(2200, 5000));
}
@Test
public void test10() {
assertEquals("[[6237, 45024100], [9799, 96079204], [9855, 113635600]]",
SumSquaredDivisors.listSquared(5000, 10000));
}
@Test
public void testA() {
System.out.println("****** Random test ******");
Random rnd = new Random();
for (int i = 0; i < 20; i++) {
int m = Math.round(rnd.nextInt(25) * (42 - 2) + 200);
int n = Math.round(rnd.nextInt(125) * (42 - 2) + 1100);
System.out.println("listSquared m: " + m + " n: " + n);
assertEquals(SumSquaredDivisorsTest.solution(m, n), SumSquaredDivisors.listSquared(m, n));
}
}
}个人总结：
import java.util.stream.LongStream;
public class SumSquaredDivisors {
public static String listSquared(long m, long n) {
// your code
String result = "[";
for (long i = m; i <= n; i++) {
if (Math.round(Math.sqrt(sumOfSquareDivisors(i))) == Math.sqrt(sumOfSquareDivisors(i))) {
result += "[" + i + ", " + sumOfSquareDivisors(i) + "], ";
}
}
return result.length() > 1 ? result.substring(0, result.length() - 2) + "]" : "[]";
}
public static long sumOfSquareDivisors(long n) {
return LongStream.range(1, n + 1).filter(i -> n % i == 0).map(i -> i * i).sum();
}
}import java.util.ArrayList;
import java.util.List;
public class SumSquaredDivisors {
public static String listSquared(long m, long n) {
List> finalRes = new ArrayList>();
for (long i = m; i <= n; i++) {
double result = 0;
for (int y = 1; y <= i; y++) {
if (i % y == 0) {
result = result + Math.pow(y, 2);
}
}
if (Math.sqrt(result) % 1 == 0) {
List pair = new ArrayList();
pair.add((int) i);
pair.add((int) result);
finalRes.add(pair);
}
}
return finalRes.toString();
}
} import java.util.ArrayList;
import java.util.stream.LongStream;
public class SumSquaredDivisors {
public static String listSquared(long m, long n) {
ArrayList strings = new ArrayList<>();
LongStream.rangeClosed(m, n).forEach(current -> {
long sum = LongStream.rangeClosed(1, current).filter(i -> current % i == 0).map(i -> i * i).sum();
if (Math.sqrt(sum) % 1 == 0)
strings.add(String.format("[%d, %d]", current, sum));
});
return String.valueOf(strings);
}
} import java.util.Arrays;
import java.util.stream.LongStream;
public class SumSquaredDivisors {
public static String listSquared(long m, long n) {
// your code
String[] tempResult = LongStream.rangeClosed(m, n).boxed().map(l -> new long[] { l, sumSquaredDivisors(l) })
.filter(ls -> isSquared(ls[1])).map(Arrays::toString).toArray(String[]::new);
return Arrays.toString(tempResult);
}
private static boolean isSquared(long n) {
long sqrtN = (long) Math.sqrt(n);
return sqrtN * sqrtN == n;
}
private static long sumSquaredDivisors(long n) {
return LongStream.rangeClosed(1, (long) Math.sqrt(n)).filter(l -> n % l == 0)
.map(l -> (l * l) + ((n / l) == l ? 0 : (n / l) * (n / l))).sum();
}
}