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-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 5e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 int T;
 LL n;
 PLL ans;

 int primes[] = {, , , , , , , , , , , , , , , };

 // 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 = , LL ret = , LL pcnt = , int limit = inf, int cnt = ) {
     if(ret > n || limit < cnt) return;
     LL tmp = pcnt * (cnt + );
     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 + ); // 选 primes[x]
     if(cnt) dfs(x + , ret, tmp, cnt, ); // 不选 primes[x]
 }

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