未验证 提交 2f2e9442 编写于 作者: S Sombit Bose 提交者: GitHub

Create Problem12.java

Added solution for problem 12 of Project Euler
上级 d0962bc9
/*
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:
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_Highly_Divisible_Triangular_Number {
/* 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) {
int sum = 0;
for (int i = 0; i <= n; i++)
sum += i;
return sum;
}
public static void main(String[] args) {
long start = System.currentTimeMillis(); // start the stopwatch
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) {
// 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");
}
}
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