Interviewstreet 的示例测试用例：方程
于是就有了一个名为 Interviewstreet.com 的网站。在这里我们可以找到具有挑战性的编程问题。不幸的是，您必须登录才能查看问题。
以下是我试图解决的问题的简要描述：
求方程组的正积分解的个数
(1/x) + (1/y) = 1/N!（读1乘n阶乘）打印一个整数，即以 1000007 为模的正积分解的数量。
例如，当 N=3 时，(x,y) 可以是：(7,42)， (9,18)、(8,24)、(12,12)、(42,7) >，<代码>(18,9)，<代码>(24,8)。或者说我是这么想的。
请帮助我，特别是已经解决了这个问题的你。我刚刚为问题方程编写了代码。我的算法有问题，我可以要求输出前 10 个整数吗？即 N=2、N=3、N=4 ... N=10 这样我就可以找出我的算法中的缺陷。谢谢:)
编辑：哦，请不要发布解决方案代码，因为它会破坏我和试图解决这个问题的人的乐趣:)
So there is a website named interviewstreet.com. Here we can find challenging programming problems. Unfortunately you have to be logged in to see the questions.
Here's a brief description of the problem I'm attempting to solve:
Find the no of positive integral solutions for the equations
(1/x) + (1/y) = 1/N!(read 1 by n factorial) Print a single integer which is the no of positive integral solutions modulo 1000007.
For example, when N=3, (x,y) can be: (7,42), (9,18), (8,24), (12,12), (42,7), (18,9), (24,8). Or so I thought.
Help me please, especially you who have solved this problem. I have just coded for the problem Equations. There is something wrong with my algorithm, can I ask for output for the first 10 integers? i.e. N=2, N=3, N=4 ... N=10 so that I can find out the flaw in my algorithm. Thanks :)
EDIT: Oh, please don't post solution code as it will ruin the fun for me and for people trying to solve this :)
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我的解决方案被采访街接受了。
首先，我的解决方案没有被接受，但是在看到@Reinardus Surya Pradhit 的帖子后，我意识到，如果对 (x, y) 会被计算两次，所以我稍微改变了它并获得了成功
我不会在这里发布我的解决方案，但我可以告诉您 N = 3 -> 中所有变量的测试用例数 = 10
这是结果
我的提示是：尝试表达N！素数如
p1^q1 * p2^q2 * ... * pn^qnMy solutions was accepted by interview street.
Firstly, my solutions wasn't accepted, but after saw @Reinardus Surya Pradhit post, i realized, if pair (x, y) will be count twice, so i change it a litter bit and got success
I will not post my solution here, but i can tell you the test case for all variable from N = 3 -> N = 10
Here the result
My hint is: try to express N! in primes from like
p1^q1 * p2^q2 * ... * pn^qn暂时忽略
N!的特殊形式，求解方程时写为
x = k + d。然后确定解决方案数量的任务留给读者作为练习。
Disregarding the special form of
N!for the moment, to solve the equationwrite
x = k + d. ThenThe task of determining the number of solutions from that is left as an exercise for the reader.
重要的是只处理整数以避免舍入错误：首先将方程重新排列为：
我不确定从哪里开始。
It is important to only deal with integers to avoid rounding errors: start by rearranging the equation to:
I'm not sure where to go from there.
为了得到最终结果，我们需要计算 (2*q1+1)*(2*q2+1)*(2*q3+1)...
但是我们将如何存储结果，假设 N=32327 将会溢出到结果之上。
如果我错了请纠正我
to arrive at final result we need to calculate (2*q1+1)*(2*q2+1)*(2*q3+1)...
But how we will store the result , let say N=32327 which will overflow above result.
Please correct me if I'm wrong