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.

 

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).

 

Output

For each test case, output an integer denoting the answer.

 

Sample Input

910 210 310 410 510 610 710 810 9100 13

 

Sample Output

121000004

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：

#include<cstdio>#include<string>#include<cstring>#include<algorithm>#include<climits>#include<map>#include<iostream>#include<vector>#include<queue>#include<stack>#include<cmath>using namespace std;int a[100005];int main(){    for(int i = 2; i <= 1000; i++)    {        if(!a[i])            for(int j = i*2; j <= 100000; j+=i)            {                a[j] = 1;            }    }    int T;    scanf("%d", &T);    while(T--)    {        int n, d, sum = 0;        scanf("%d%d", &n, &d);        for(int i = 2; i <= d; i++)        {            if(i*d >= n) break;            if(!a[i]) sum++;            if(!a[i] && d%i==0 && d!=i) break;        }        printf("%d\n", sum);    }}