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：
1 #pragma GCC optimize("Ofast") 2 #include <bits/stdc++.h> 3 using namespace std; 4 5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 6 #define Rep(i,n) for (int i = 0; i < (n); ++i) 7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 13 14 #define pr(x) cout << #x << " = " << x << " " 15 #define prln(x) cout << #x << " = " << x << endl 16 17 #define LOWBIT(x) ((x)&(-x)) 18 19 #define ALL(x) x.begin(),x.end() 20 #define INS(x) inserter(x,x.begin()) 21 22 #define ms0(a) memset(a,0,sizeof(a)) 23 #define msI(a) memset(a,inf,sizeof(a)) 24 #define msM(a) memset(a,-1,sizeof(a)) 25 26 #define MP make_pair 27 #define PB push_back 28 #define ft first 29 #define sd second 30 31 template<typename T1, typename T2> 32 istream &operator>>(istream &in, pair<T1, T2> &p) { 33 in >> p.first >> p.second; 34 return in; 35 } 36 37 template<typename T> 38 istream &operator>>(istream &in, vector<T> &v) { 39 for (auto &x: v) 40 in >> x; 41 return in; 42 } 43 44 template<typename T1, typename T2> 45 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 46 out << "[" << p.first << ", " << p.second << "]" << "\n"; 47 return out; 48 } 49 50 typedef long long LL; 51 typedef unsigned long long uLL; 52 typedef pair< double, double > PDD; 53 typedef pair< int, int > PII; 54 typedef pair< LL, LL > PLL; 55 typedef set< int > SI; 56 typedef vector< int > VI; 57 typedef map< int, int > MII; 58 typedef vector< LL > VL; 59 typedef vector< VL > VVL; 60 const double EPS = 1e-10; 61 const int inf = 1e9 + 9; 62 const LL mod = 1e9 + 7; 63 const int maxN = 5e5 + 7; 64 const LL ONE = 1; 65 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 66 const LL oddBits = 0x5555555555555555; 67 68 int T; 69 LL n; 70 PLL ans; 71 72 int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53}; 73 74 // x 表示当前处理到第 x 个质数 75 // ret为当前选择下的质数乘积 76 // pcnt 为 1~x-1 个质数中，每个质数选择数量+1的乘积 77 // limit 表示第x个质数选则的上限 78 // cnt表示第 x 个质数已经选了多少个 79 inline void dfs(int x = 0, LL ret = 1, LL pcnt = 1, int limit = inf, int cnt = 0) { 80 if(ret > n || limit < cnt) return; 81 LL tmp = pcnt * (cnt + 1); 82 if(ans.sd < tmp || ans.sd == tmp && ans.ft > ret) ans = MP(ret, tmp); 83 84 if(n / ret >= primes[x])dfs(x, ret * primes[x], pcnt, limit, cnt + 1); // 选 primes[x] 85 if(cnt) dfs(x + 1, ret, tmp, cnt, 0); // 不选 primes[x] 86 } 87 88 int main(){ 89 INIT(); 90 cin >> T; 91 while(T--) { 92 cin >> n; 93 ans = MP(inf, -1); 94 dfs(); 95 cout << ans.ft << " " << ans.sd << endl; 96 } 97 return 0; 98 }