Problem 12

问题描述：
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?


思路：
以28为例子，
28的因子为1,2,4,7,14,28
28的质因子为2和7
28=2^2*7
2出现了两次
7出现了依次
根据排列组合
28的因子个数为(2+1)*(1+1)

任意数N，其质因子为a1,a2,...,an
其所有的因子都可以表示为a1^b1 * a2^b2 * ... *an^bn

实现：

public static int count_divisors(long number) {
		int count = 0;
		//先找到因子和因子的最高次方
		int[] divisors = new int[500];
		int index = 0;
		int divis = 3;
		Arrays.fill(divisors, 0);
		if(number%2==0){
			do{
				number=number/2;
				count++;
			}while(number%2==0);
			divisors[index] = count;
			index++;
		}
		while(number>1){
			count = 0;
			while(number%divis==0){
					number = number/divis;
					count++;
			}
			if(count!=0){
				divisors[index] = count;
				index++;				
			}
			divis+=2;
		}
		count = 1;
		for(int i=0; divisors[i]>0;i++){
		
			count*=(divisors[i]+1);
		}
		
		return count;
	}
	
	public static int find_number(){
		int number = 28;
		int count = 6;
		int step = 8;
		while(count<500){
			number = number + step;
			step++;
			count = count_divisors(number);
			
		}
		return number;
	}