Problem 12:Highly divisible triangular number

原题链接:http://projecteuler.net/problem=12


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?


题目大意是:

三角形数序列是由对自然数的连加构造而成的。所以第七个三角形数是1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. 那么三角形数序列中的前十个是:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

下面我们列出前七个三角形数的约数:

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

可以看出28是第一个拥有超过5个约数的三角形数。

那么第一个拥有超过500个约数的三角形数是多少?


解法1:

从第一个三角形数开始,计算每个三角形数的约数个数,直到某个三角形数具有超过500个约数为止。而计算约数个数的方法是:先初始化个数为2,即1和它本身;然后从i=2开始,到这个数的平方根的向下取整,能被i整除则个数加2;如果这个数是个平方数,我们需要将个数减去1。

php代码如下所示:

$over_num = 500;
$count_of_divisors = 0;
$sum = 0;
$i = 1;

function get_divisors_num($number){
	$sqrt_root = floor(sqrt($number));
	$num = 2;
	for($i=2;$i<=$sqrt_root;$i++){
		if($number%$i==0){
			$num+=2;
		}
	}
	if(pow($sqrt_root,2)==$number){
		$num -=1;
	}

	return $num;
}


while($count_of_divisors<=$over_num){
	$sum += $i;
	$i++;
	$count_of_divisors = get_divisors_num($sum);
}

echo $sum,"\n";


注:题目的中文翻译源自http://pe.spiritzhang.com

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