HDU 5750 Dertouzos

Dertouzos

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1643    Accepted Submission(s): 514

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

 

Source

BestCoder Round #84

 

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官方题解：

随便推导下, 令$y=xd$, 如果$d$是$y$的maximum positive proper divisor, 显然要求$x$是$y$的最小质因子. 令$mp(n)$表示$n$的最小质因子, 那么就有$x\le mp(d)$, 同时有$y<n$, 那么x \le \lfloor \frac{n-1}{d} \rfloorx≤⌊​d​​n−1​​⌋. 于是就是计算有多少个素数$x$满足x \le \min{mp(d), \lfloor \frac{n-1}{d} \rfloor}x≤min{mp(d),⌊​d​​n−1​​⌋}.

当$d$比较大的时候, \lfloor \frac{n-1}{d} \rfloor⌊​d​​n−1​​⌋比较小, 暴力枚举$x$即可. 当$d$比较小的时候, 可以直接预处理出答案. 阈值设置到10^6 \sim 10^710​6​​∼10​7​​都可以过.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int prime[(int)1e5+5];void getP(int MAXN){    memset(prime,0,sizeof(prime));    for(int i=2; i<=MAXN; i++)    {        if(!prime[i])prime[++prime[0]]=i;        for(int j=1; j<=prime[0]&&prime[j]<=MAXN/i; j++)        {            prime[prime[j]*i]=1;            if(i%prime[j]==0) break;        }    }}int main(){    getP((int)1e5+5);    int t;    scanf("%d",&t);    while(t--){        int n,d;        scanf("%d%d",&n,&d);        int ans=0;        for(int i=1;i<=prime[0];i++){            if(prime[i]*d>=n) break;//枚举质因子，若质因子×d>=上限说明已统计完毕            ++ans;            if(d%prime[i]==0) break;//考虑到maximum positive proper divisor也是合数的情况        }        printf("%d\n",ans);    }    return 0;}