hud 5750 Dertouzos (数论+暴力)

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

参考博客： 点击打开链接 这里的思路说的很明确。从他的博客里学到了很多

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
#include<utility>
#include<iomanip>
#include<time.h>
typedef long long ll;
const double Pi = acos(-1.0);
const int N = 1e6+10, M = 1e3+20, mod = 1e9+7, inf = 2e9+10;
const double e=2.718281828459 ;
const double esp=1e-9;
using namespace std;
int t;
bool flag[N];
int prime[N];
int num=0;
void Prim()
{
    memset(flag,true,sizeof(flag));
    for(int i=2; i<100000; i++)
        if(flag[i])
        {
            prime[num++]=i;
            for(int j=2*i; j<100000; j+=i)
                flag[j]=false;
        }
}
int factor(int d,int n)
{
    int sum=0;
    for(int i=0; i<num; i++)
    {
        if(d*prime[i]>=n) break;
        if(d>=prime[i])
        sum++;
        if(d%prime[i]==0) break;//如果d被prime[i]整除,
        //这说明d中含有素数且小于d，所以再往下都不会出现最大因子是d,
        //而是(d/prime[i](这个prime[i]整除d))*prime[i]
    }
    return sum;
}
int main()
{
    Prim();
    while(~scanf("%d",&t))
    {
        for(int i=0; i<t; i++)
        {
            int n,d;
            scanf("%d%d",&n,&d);
            int g=factor(d,n);
            printf("%d\n",g);
        }
    }
    return 0;
}