hdu-5750 Dertouzos(数论)

题目链接：

Dertouzos

Time Limit: 7000/3500 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, output an integer denoting the answer.

[align=left]Sample Input[/align]

9

10 2

10 3
10 4

10 5
10 6

10 7

10 8

10 9

100 13

[align=left]Sample Output[/align]

1
2

1

0

0

0

0

0

4

题意：

就是给一个n和一个d,问有多少个小于n的数的最大因子是d;

思路：

个数为min((n-1)/d,d')d'为d的最小质因子;
素数筛,然后枚举最小质因子,当时忘加一个条件最后测的时候t了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-8;

int prime
,sum
,a
,cnt=0,n,d;
void Init()
{
sum[1]=0;
For(i,2,N-maxn)
{
if(!prime[i])
{
for(int j=2*i;j<N-maxn;j+=i)
{
prime[j]=1;
}
sum[i]=sum[i-1]+1;
}
else sum[i]=sum[i-1];
}
For(i,2,N-maxn)
{
if(!prime[i])a[++cnt]=i;
}
}

inline int check(int x)
{
for(int i=1;i<=cnt;i++)
{
if(x%a[i]==0)return a[i];
if((LL)a[i]*a[i]>x||a[i]>n/d)break;
}
return x;
}
int main()
{
int t;
read(t);
Init();
while(t--)
{
read(n);read(d);
n--;
int le=min(check(d),n/d);
printf("%d\n",sum[le]);
}

return 0;
}