微软笔试-Divisors

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

Given an integer n, for all integers not larger than n, find the integer with the most divisors. If there is more than one integer with the same number of divisors, print the minimum one.

输入

One line with an integer n.
For 30% of the data, n ≤ 103
For 100% of the data, n ≤ 1016

输出

One line with an integer that is the answer.

样例输入

100

样例输出

60

代码区：

// MS.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <math.h>
#include <set>
#include <algorithm>
using namespace std;
int primes[14]{ 2,3,5,7,11,13,17,19,23,29,31,37,41,43};
int maxdivs;
long long result;
long long N;
void DFS(long long now,int divs, int primeIndex, int lastni)
{
if (now > N)
return;
if (maxdivs < divs || (maxdivs == divs&&now < result))
{
maxdivs = divs;
result = now;
}
int i = 1;
while (now*pow(primes[primeIndex], i) <= N && i <= lastni)
{
long long newvalue = now*pow(primes[primeIndex],i);
DFS(newvalue, divs*(i + 1), primeIndex + 1, i);
i++;
}
}
int main()
{
while (cin >> N)
{
int maxni = log2(N);
result = 1;
maxdivs = 0;
for (int i = 0; i <= maxni; ++i)
{
DFS(pow(2, i), i + 1,1, i);
}
cout << result << endl;
}
return 0;
}