ZOJ2562：More Divisors（反素数）

最新推荐文章于 2018-07-13 11:07:00 发布

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本文介绍了一种算法，用于寻找不超过给定数值n的最大因子数量的整数，并提供了完整的C++实现代码。该问题适用于数学竞赛及算法挑战，通过递归深度优先搜索策略，结合质数分解技巧，有效地解决了寻找具有最多因子的整数的问题。

More Divisors

Time Limit: 2 Seconds Memory Limit: 65536 KB

Everybody knows that we use decimal notation, i.e. the base of our notation is 10. Historians say that it is so because men have ten fingers. Maybe they are right. However, this is often not very convenient, ten has only four divisors -- 1, 2, 5 and 10. Thus, fractions like 1/3, 1/4 or 1/6 have inconvenient decimal representation. In this sense the notation with base 12, 24, or even 60 would be much more convenient.

The main reason for it is that the number of divisors of these numbers is much greater -- 6, 8 and 12 respectively. A good quiestion is: what is the number not exceeding n that has the greatest possible number of divisors? This is the question you have to answer.

Input:

The input consists of several test cases, each test case contains a integer n (1 <= n <= 10¹⁶).

Output:

For each test case, output positive integer number that does not exceed n and has the greatest possible number of divisors in a line. If there are several such numbers, output the smallest one.

Sample Input:

10
20
100

Sample Output:

6
12
60

题意：求n以内因子最多的数

# include <iostream>
# define ULL unsigned long long

using namespace std;
int best, p[16] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};  ;
ULL ans, n;

void dfs(int depth, ULL tmp, int num)
{
    if(depth >= 16) return;
    if(num > best)
    {
        best = num;
        ans = tmp;
    }
    else if(num == best && tmp < ans) ans = tmp;
    for(int i=1; i<=64; ++i)
    {
        if(n / p[depth] < tmp || num > best) break;
        dfs(depth+1, tmp *= p[depth], num*(i+1));
    }
}

int main()
{

    while(cin >> n)
    {
        ans = ~0ULL;
        best = 0;
        dfs(0, 1, 1);
        cout << ans << endl;
    }
    return 0;
}