HDU 5750 bestcoder Round #84 dertouzos (Prime number screening) __hdu

Dertouzos Time limit:7000/3500 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 1518 accepted submission (s): 484
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 are 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 are d. Input There are multiple Test cases. The "a" of input contains an integer T (1≤t≤106), indicating the number of test cases. For each test case:

The contains two integers n and D (2≤n,d≤109). Output for each test case, output an integer denoting the answer. Sample Input

9 2 3 Ten 4 5 a 6 a 7 a 8 9

1 2 1 0 0 0 0 0 4 Source bestcoder Round #84 recommend wange2014 | We have carefully selected several similar problems for you:5751 5746 5745 5744 5743: give you t N and D, and let you ask for the largest divisors in the number less than n (no including itself) is the number of D.

Because to make the maximum factor D, there must be x*d=m, and X must be a prime number, and x must be less than D's minimum decomposition, so find n before a few m set up. Because T range is very large, so you can first play the table to select the list of quality and then linear sweep it.

AC Code:

#include <bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include < algorithm>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace Std;

int isprime[100005]={0};
int prime[100005]={0};
int k=0;

int init ()
{
	int i,j;
	
	for (i=2;i<100000;i++)
	{
		if (isprime[i]==0)
		{for
			(j=2;i*j<100000;j++)
				isprime[i*j ]=1;
			prime[k]=i;
			k++
		}
	}
	return 0;
}

int main ()
{
	int t,i;
	int m,n;	
	scanf ("%d", &t);
	Init ();
	while (t--)
	{
		scanf ("%d%d", &m,&n);

		for (i=0;i<k;i++)
		{
			if (n*prime[i]>=m) break;
			if (N<prime[i]) break;
			if (n%prime[i]==0) break;
		if (N*prime[i]>=m | | n<prime[i]) i--;
		printf ("%d\n", i+1);
	}
	return 0;
}