思路题 素因子 HDU 5750

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5750

Dertouzos

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 2172    Accepted Submission(s): 672

   Problem Description

A positive proper divisor is a positive divisor of a number nn, excluding nn itself. For example, 1, 2, and 3 are positive proper divisors of 6, but 6 itself is not. 

Peter has two positive integers nn and dd. He would like to know the number of integers below nn whose maximum positive proper divisor is dd.

Input

There are multiple test cases. The first line of input contains an integer TT (1≤T≤106)(1≤T≤106), indicating the number of test cases. For each test case: 

The first line contains two integers nn and dd (2≤n,d≤109)(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

题意：为你一个n和一个d；在2到n中找出一些数，例如6，它有因子2，3（因子除了1和它本身，下同）;所以6的最大因子是3，

现在在2到n中找出一些数，这些数的最大因子是d；求这些数有多少个；

思路：假设一个数为a，使得d*a<n；如果a不是素数的话，a总是有一个因子b，使得b*d>d，即a*d的因子b*d比因子d大，所以a*d一定不符合条件，如果a>d&&d*a<n，a*d也一定不符合条件,因为a比d大;如果a比d的其中一个因子大，就可以替换那个因子得到比d大的数字，所以当d%a==0；时说明比a大的素数一定不行，直接break；

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<map>#include<string>#define LL long long int#define inf 0x3f3f3f3f#define N 100010#define mod 10000007using namespace std;bool a[50100]={0};int b[10000];int main(){    int sum=0;    for(int i=2;i<=50000;i++)    {        if(!a[i])        {            for(int j=i*2;j<=50000;j+=i)                a[j]=1;            b[sum++]=i;        }    }    int t;    scanf("%d",&t);    while(t--)    {        int n,d;        int ans=0;        scanf("%d%d",&n,&d);        for(int i=0;b[i]*d<n&&b[i]<=d;i++)        {            ans++;            if(d%b[i]==0)                break;        }            printf("%d\n",ans);    }}