ural 1748. The Most Complex Number

1748. The Most Complex Number

Time limit: 1.0 second
Memory limit: 64 MB

Let us define a complexity of an integer as the number of its divisors. Your task is to find the most complex integer in range from 1 to n. If there are many such integers, you should find the minimal one.

Input

The first line contains the number of testcases t (1 ≤ t ≤ 100). The i-th of the following t lines contains one integer n_i (1 ≤ n_i ≤ 10¹⁸).

Output

For each testcase output the answer on a separate line. The i-th line should contain the most complex integer in range from 1 to n_i and its complexity, separated with space.

Sample

input	output
5 1 10 100 1000 10000	1 1 6 4 60 12 840 32 7560 64

Problem Author: Petr Lezhankin
Problem Source: Ufa SATU Contest. Petrozavodsk Summer Session, August 2009

题目大意：一个数的因子的个数为它的复杂度，求[1, n]间因子最多的数， 以及该数因子的个数

思路：有个公式。。

N = p1^a1*p2^a2*p3^a3*p4^a4，有a1>= a2 >= a3 >= a4， p1,p2, p3, p4为质数

number = (1+a1)*(1+a2)*(1+a3)*(1+a4)

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>
#define FOR(i, a, b)  for(int i = (a); i <= (b); i++)
#define RE(i, n) FOR(i, 1, n)
#define FORP(i, a, b) for(int i = (a); i >= (b); i--)
#define REP(i, n) for(int i = 0; i <(n); ++i)
#define SZ(x) ((int)(x).size )
#define ALL(x) (x).begin(), (x.end())
#define MSET(a, x) memset(a, x, sizeof(a))
using namespace std;


typedef unsigned long long ll;
typedef pair<int, int> P;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const double pi=3.14159265358979323846264338327950288L;
const double eps=1e-6;
const int mod = 1e9 + 7;
const ll INF = ~0LL;
const int MAXN = 1005;
const int xi[] = {0, 0, 1, -1};
const int yi[] = {1, -1, 0, 0};

ll prime[] = {1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
ll n, res1, res2;
void dfs(ll now, ll cnt, ll limt, int pos){

    if(now > n) return;
    if(pos > 15) return;
    if(cnt > res1) res1 = cnt, res2 = now;
    else if(cnt == res1 && res2 > now) res2 = now;

    for(ll i = 1; i <= limt; i++){
            if((double)(now)*(prime[pos]) > double(n)) break;
        dfs(now *= prime[pos], cnt*(1+i), i, pos+1);
    }
}
int main() {
    //freopen("in.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while(t--){
        res1 = 0, res2 = INF;
        cin >> n;
        dfs(1, 1, 60, 1);
        cout << res2 <<" " << res1 << endl;
    }
    return 0;
}