Dertouzos
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 577 Accepted Submission(s): 160
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
思路:
题目要求的是，在小于n的范围内，找出所有最大除数为d的数。假设满足该条件的一个数y，那么y=x*d，x为y的最小质因子。因为x已经无法再分解为2个大于1的数 x=a*b，使得 y=a*（b*d）。 接下来再看，另slove（n）表示n的最小质因子。如果d是非质数 那么d=slove（d）*k，y=x*k*slove（d）并且x*k<=d，由此可得x<=solve（d），因为若x>solve（d），那么x*k>d，d就不是y的最大除数了；如果d是质数，那么solve（d）=d，也满足以上结论。
同时题目还限定范围小于n，那么可以得出x<=（n-1）/d。综上所述，取2种情况的并集，那么就是 x<=min（solve(d)，(n-1)/d）并且x是质数。并且由于
（n-1）/d，那么x最大不会超过sqrt（n），所以先把sqrt（1e9）内的素数先筛选 用数组存起来，然后再暴力枚举x就可以了。
#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <sstream>
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define mod 1000000007
#define INF 1000000006
using namespace std;
typedef long long LL;
int n,d,k,p;
vector<LL> Q;
bool vis[+];
void init()
{
for(LL i=;i<=+;i++)
{
if(!vis[i])
{
Q.push_back(i);
for(LL j=i*i;j<=+;j+=i)
{
vis[j]=true;
}
}
}
}
int main()
{
#ifdef Local
freopen("data.txt","r",stdin);
#endif
int T,i,j,h,r,sum=,m,ans,flag,q,mid,l;
cin>>T;
init();
while(T--)
{
ans=flag=;
scanf("%d%d",&n,&d);
n=(n-)/d;
for(i=;Q[i]<=n&&i<Q.size();i++)
{
ans++;
if((d%Q[i])==)break;
}
cout<<ans<<endl;
}
}