BestCoder Round #84 1004Dertouzos

最新推荐文章于 2016-08-01 17:52:26 发布

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本文介绍了一种算法，用于解决求解指定范围内整数的最大非平凡因子为特定值的问题。通过分解因子并利用枚举技巧，文章提供了一个有效的解决方案。

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

Input

There are multiple test cases. The first line of input contains an integer

(1≤T≤106), indicating the number of test cases. For each test case:

The first line contains two integers

n and

(2≤n,d≤109).

Output

For each test case, output an integer denoting the answer.

Sample Input

Sample Output

题意：给你到n的数列，求这里边的数最大非平凡因子（非平凡：不等于1和它本身）为d的个数。

思路：a*d=n（d为n的最大非平凡因子，d<n&&a>1）d=p*p'（p为d的最大非平凡因子，p<d&&p'>1）==> n=a*p*p' 范围：1<a<=p'<=p；根据该范围枚举每一个因子。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>

using namespace std;

int is[100000 +5];

int ds()
{
    is[0] = 2;
    is[1] = 3;
    int k = 2;
    for(int i = 5; i <= 100000; i++)
    {
        int temp = sqrt(i);
        bool judge = 0;
        for(int j = 2; j <= temp; j++)
            if(i%j == 0)
            {
                judge = 1;
                break;
            }

        if(judge == 0)
            is[k++] = i;
    }
    return k;
}
int main()
{
    int k = ds();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,d,sum = 0;
        scanf("%d%d",&n,&d);
        //int temp = sqrt(d);
        for(int i = 0; is[i] <= d; i++)
        {
            if(is[i]*d < n)
                sum++;
            else
                break;
            if(d%is[i] == 0)          //a<p'
                break;

        }
        printf("%d\n",sum);
    }

}