URAL 1748 The Most Complex Number

题目链接：https://vjudge.net/problem/11177

题目大意：

　　求小于等于 n 的最大反素数。

分析：

　　n <= 10^18，而前20个素数的乘积早超过10^18，因此可手动打素数表，再dfs寻找最大反素数。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "\n";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 5e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int T;
LL n;
PLL ans;

int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

// x 表示当前处理到第 x 个质数
// ret为当前选择下的质数乘积 
// pcnt 为 1~x-1 个质数中，每个质数选择数量+1的乘积 
// limit 表示第x个质数选则的上限 
// cnt表示第 x 个质数已经选了多少个 
inline void dfs(int x = 0, LL ret = 1, LL pcnt = 1, int limit = inf, int cnt = 0) {
    if(ret > n || limit < cnt) return;
    LL tmp = pcnt * (cnt + 1);
    if(ans.sd < tmp || ans.sd == tmp && ans.ft > ret) ans = MP(ret, tmp);
    
    if(n / ret >= primes[x])dfs(x, ret * primes[x], pcnt, limit, cnt + 1); // 选 primes[x]
    if(cnt) dfs(x + 1, ret, tmp, cnt, 0); // 不选 primes[x]
}

int main(){
    INIT();
    cin >> T;
    while(T--) {
        cin >> n;
        ans = MP(inf, -1);
        dfs();
        cout << ans.ft << " " << ans.sd << endl;
    }
    return 0;
}