HDU5750(数论，素数筛法)
Dertouzos
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2125 Accepted Submission(s): 647
Problem Description
A positive proper divisor is a positive divisor of a number
n, excluding
n itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not.
Peter has two positive integers n and d. He would like to know the number of integers below n whose maximum positive proper divisor is d.
Peter has two positive integers n and d. He would like to know the number of integers below n whose maximum positive proper divisor is d.
Input
There are multiple test cases. The first line of input contains an integer
T
(1≤T≤106), indicating the number of test cases. For each test case:
The first line contains two integers n and d (2≤n,d≤109).
The first line contains two integers n and d (2≤n,d≤109).
Output
For each test case, output an integer denoting the answer.
Sample Input
9 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 100 13
Sample Output
1 2 1 0 0 0 0 0 4
题解：题目大意：老规矩，去读读吧！
我的思路：先筛一遍素数，筛素数的时候只需要筛选到1e9开根号。至于为什么，因为他是要找小于n的切以d为最大因子的数。那么最多另外一个因子等于d，不可能大于d，因为大于d，d就不是最大因子了也就是d的平方小于n，所以素数筛出1e5就行了。然后从2到d的做小素因子跑一遍，素数的个数就最终的结果。这一点自己可以去思考一下。（为什么一定要是素数才满足题意？ 还有就是为什么只需要遍历到d的最小素因子就行了？自己举个反例思考一下，对于搞算法的应该不难吧！）
下面附上代码：
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