URAL 1748 The Most Complex Number

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 << "]" << "
";
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 }
View Code