题目链接:https://vjudge.net/problem/11177
求小于等于 n 的最大反素数。
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 // 欧拉筛法 73 const int N = 100; 74 bool isPrime[N + 7];//isPrime[i]表示i是不是质数 75 VI primes;//primes用来存质数 76 inline void Euler(){ 77 For(i, 2, N) isPrime[i] = true;//初始化为质数 78 For(i, 2, N) { 79 if(isPrime[i]) primes.PB(i);//把质数存起来 80 for(int j = 0; j < primes.size() && i * primes[j] <= N; ++j) { 81 isPrime[i * primes[j]] = false; 82 if(i % primes[j] == 0) break;//保证每个合数被它最小的质因数筛去 83 } 84 } 85 } 86 87 // x 表示当前处理到第 x 个质数 88 // ret为当前选择下的质数乘积 89 // fcnt 为 1~x-1 个质数中,每个质数选择数量+1的乘积 90 // prev_cnt表示第 x-1 个质数已经选了多少个 91 // cnt表示第 x 个质数已经选了多少个 92 inline void dfs(int x, LL ret, LL fcnt, int prev_cnt, int cnt) { 93 if(ret > n || prev_cnt < cnt) return; 94 95 if(ans.sd < fcnt * (cnt + 1)) ans = MP(ret, fcnt * (cnt + 1)); 96 else if(ans.sd == fcnt * (cnt + 1) && ans.ft > ret) ans = MP(ret, fcnt * (cnt + 1)); 97 98 if(n / ret >= primes[x])dfs(x, ret * primes[x], fcnt, prev_cnt, cnt + 1); // 选 primes[x] 99 if(cnt) dfs(x + 1, ret, fcnt * (cnt + 1), cnt, 0); // 不选 primes[x] 100 } 101 102 int main(){ 103 INIT(); 104 cin >> T; 105 Euler(); // 预先求出100以内的素数 106 while(T--) { 107 cin >> n; 108 ans = MP(inf, -1); 109 dfs(0, 1, 1, inf, 0); 110 cout << ans.ft << " " << ans.sd << endl; 111 } 112 return 0; 113 }
URAL 1748 The Most Complex Number
原文:https://www.cnblogs.com/zaq19970105/p/10800152.html