HDU 5750 BestCoder Round #84 Dertouzos （素数筛选）

Dertouzos

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

Total Submission(s): 1518    Accepted Submission(s): 484

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

9

10 2

10 3

10 4

10 5

10 6

10 7

10 8

10 9

100 13

Sample Output

1

2

1

4

Source

​​BestCoder Round #84​​

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题解：给你 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。

       因为要使最大因数为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成立即可。因为 t 范围很大，所以可以先打表筛选出质数表再线性扫一遍就可以了。

AC代码：

//#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;

int isprime[100005]={0};
int prime[100005]={0};
int k=0;

int init()
{
  int i,j;
  
  for(i=2;i<100000;i++)
  {
    if(isprime[i]==0)
    {
      for(j=2;i*j<100000;j++)
        isprime[i*j]=1;
      prime[k]=i;
      k++;
    }
  }
  return 0;
}

int main()
{
  int t,i;
  int m,n;  
  scanf("%d",&t);
  init();
  while(t--)
  {
    scanf("%d%d",&m,&n);

    for(i=0;i<k;i++)
    {
      if(n*prime[i]>=m) break;
      if(n<prime[i]) break;
      if(n%prime[i]==0) break;
    }
    if(n*prime[i]>=m || n<prime[i]) i--;
    printf("%d\n",i+1);
  }
  return 0;
}