HDU 5750

传送门：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): 871    Accepted Submission(s): 268

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

题意：给出t个n和d，让你求小于n的数中最大约数（不包括本身）为d的数量

题解：要使最大因数为d，必存在x*d=m，同时x必须为质数，且x必须小于d的最小质因数，因此找n前面有几个m即可

因为t范围很大，所以可以先筛选出质数表再线性扫一遍

恩，果然数论超级差系列

问题数论差就算了做题还有谜之自信这个方法肯定对...要检讨一下了

下面代码

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<stdlib.h>#include<math.h>#include<map>#include<vector>#include<set>#include<queue>using namespace std;int isprime[100005]={0};int primelist[100005]={0};int k=0;int prime(){int i,j;for(i=2;i<100000;i++){if(isprime[i]==0){for(j=2;i*j<100000;j++)isprime[i*j]=1;primelist[k]=i;k++;}}return 0;}int main(){int t,i;int m,n;scanf("%d",&t);prime();while(t--){cin>>m>>n;for(i=0;i<k;i++){if(n*primelist[i]>=m) break;if(n<primelist[i]) break;if(n%primelist[i]==0) break;}if(n*primelist[i]>=m || n<primelist[i]) i--;printf("%d\n",i+1);}return 0;}