hdu 5750(数论)

Dertouzos

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 891 Accepted Submission(s): 274


[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:如果 d 是k除了自身以外最大的因子, 在[2,n)内找有多少个数满足 最大的因子是 k .
题解:昨天想得太简单了,真是naive..我开始的想法是 将 min(d,(n-1)/d) 里面所有的质数算出来，然后天真的以为这就是结果...然后一直WA，忽视了一个很重要的条件,那就是我们的数中d一定是最大的因子,所以如果出现了能够将d整除的质数,那么我们就可以将d分解，从而得到一个更大的 d' ,所以说我们选择的质数是不能够将 d 除尽的.

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int N = 100005;
bool p
;
int prime
;
int id;
void init()
{
id = 0;
for(int i=2; i<N; i++)
{
if(!p[i])
{
prime[++id] = i;
for(LL j=(LL)i*i; j<N; j+=i)
{
p[j] = true;
}
}
}
}

int main()
{
init();
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
int n,d;
scanf("%d%d",&n,&d);
int ans = 0;
for(int i=1; i<=id; i++)
{
if(prime[i]*d>=n) break;
ans++;
if(d%prime[i]==0) break;
}
printf("%d\n",ans);
}
return 0;
}