hdu5750 Dertouzos （思路題）

Dertouzos

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2097    Accepted Submission(s): 634

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 0 0 0 0 0 4

 

題意：找2~n-1中有幾個數的最大因子是d

思路：一個數 d*x  的最大因子是 d  那麼 x 必是素數，且 x 小於等於 d 的最大素因子 （比賽時看樣例100 13，d恰好是素數， 只想着找到有多少個小於等於d的素數就可以了 ，沒有考慮到d不是素數時d本身就有因子，不可能是最大的因子了，如：42  d爲14時，42的最大因子是21 而不是14 ）

code：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<stack>
#include<math.h>
#include<algorithm>
using namespace std;
int n,d,k=0;
int vis[1000005],prime[1000005];
void get_prime()
{
	k=0;
    vis[0]=vis[1]=1;
    prime[0]=prime[1]=0;
    for(int i=2; i<=1000000; i++)
    {
        if(vis[i]==0)
        {
            prime[k++]=i;
            for(int j=i+i; j<=1000000; j+=i)
            {
                vis[j]=1;
            }
        }
    }
}
int main()
{
    int T;
    get_prime();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&d);
         int ans=0;
      for(int i=0;i<k;i++)
      {
      	if(prime[i]>d||prime[i]*d>=n)
      	break;
      	if(d%prime[i]==0)
      	{
      		ans++;
      		break;
      	}
      	ans++;
      }
         printf("%d\n",ans);
    }
}