reformat

realDuYuanChao · realDuYuanChao · commit 8ad87ce9d3c0 · 2020-10-06T14:58:38.000+08:00
diff --git a/ProjectEuler/Problem12.java b/ProjectEuler/Problem12.java
@@ -1,24 +1,34 @@
-/*
-    The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
-    The first ten terms would be:
+package ProjectEuler;
+/**
+ * The sequence of triangle numbers is generated by adding the natural numbers.
+ * So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
+ * The first ten terms would be:
+ * <p>
+ * 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+ * <p>
+ * Let us list the factors of the first seven triangle numbers:
+ * <p>
+ * 1: 1
+ * 3: 1,3
+ * 6: 1,2,3,6
+ * 10: 1,2,5,10
+ * 15: 1,3,5,15
+ * 21: 1,3,7,21
+ * 28: 1,2,4,7,14,28
+ * We can see that 28 is the first triangle number to have over five divisors.
+ * <p>
+ * What is the value of the first triangle number to have over five hundred divisors?
+ * <p>
+ * link: https://projecteuler.net/problem=12
+ */
+public class Problem12 {
 
-    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
-
-    Let us list the factors of the first seven triangle numbers:
-
-     1: 1
-     3: 1,3
-     6: 1,2,3,6
-    10: 1,2,5,10
-    15: 1,3,5,15
-    21: 1,3,7,21
-    28: 1,2,4,7,14,28
-    We can see that 28 is the first triangle number to have over five divisors.
-
-    What is the value of the first triangle number to have over five hundred divisors?
-*/
-
-public class Problem_12 {
+    /**
+     * Driver Code
+     */
+    public static void main(String[] args) {
+        assert solution1(500) == 76576500;
+    }
 
     /* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
     public static int triangleNumber(int n) {
@@ -28,36 +38,29 @@ public static int triangleNumber(int n) {
         return sum;
     }
 
-    public static void main(String[] args) {
-
-        long start = System.currentTimeMillis(); // start the stopwatch
-        
+    public static int solution1(int number) {
         int j = 0; // j represents the jth triangle number
         int n = 0; // n represents the triangle number corresponding to j
         int numberOfDivisors = 0; // number of divisors for triangle number n
-        
-        while (numberOfDivisors <= 500) {
+
+        while (numberOfDivisors <= number) {
 
             // resets numberOfDivisors because it's now checking a new triangle number
             // and also sets n to be the next triangle number
             numberOfDivisors = 0;
             j++;
             n = triangleNumber(j);
-            
+
             // for every number from 1 to the square root of this triangle number,
             // count the number of divisors
             for (int i = 1; i <= Math.sqrt(n); i++)
                 if (n % i == 0)
                     numberOfDivisors++;
-            
+
             // 1 to the square root of the number holds exactly half of the divisors
             // so multiply it by 2 to include the other corresponding half
             numberOfDivisors *= 2;
         }
-        
-        long finish = System.currentTimeMillis(); // stop the stopwatch
-        
-        System.out.println(n);
-        System.out.println("Time taken: " + (finish - start) + " milliseconds");
+        return n;
     }
 }
