反素数学习_The Most Complex Number

我就是我 2022-05-12 22:10 146阅读 0赞

# [题目链接][Link 1] #

先说一下反素数:(引自百度百科)

基本概念  
  
定义  
  
对于任何正整数x,其约数的个数记做g(x).例如g(1)=1,g(6)=4.如果某个正整数x满足:对于任意i(0<i<x),都有g(i)<g(x),则称x为反素数.  
性质  
  
性质一:一个反素数的质因子必然是从2开始连续的质数.  
性质二:p=2^p1\*3^p2\*5^p3\*7^p4.....必然p1>=p2>=p3>=....

这两条性质决定了搜索复杂度不会很高

想了想大概的证明:对于1-x的反素数n,可以表示为a1^p1 \* a2^p2 \* .... an^pn

1.假如a不连续,即存在ai 和 ai+1在素数表中不相邻,那么总可以把ai+1换成比ai大的最小的素数,仍然使得结果不变.得证

2.假如pi < pj (i < j),那么将ai与aj的个数对调,结果也不变,得到的数减少.得证

#include <cstdio>
    #include <ctime>
    #include <cstdlib>
    #include <cstring>
    #include <queue>
    #include <string>
    #include <set>
    #include <stack>
    #include <map>
    #include <cmath>
    #include <vector>
    #include <iostream>
    #include <algorithm>
    #include <bitset>
    #include <fstream>
    using namespace std;
    
    
    //LOOP
    #define FF(i, a, b) for(int i = (a); i < (b); ++i)
    #define FE(i, a, b) for(int i = (a); i <= (b); ++i)
    #define FED(i, b, a) for(int i = (b); i>= (a); --i)
    #define REP(i, N) for(int i = 0; i < (N); ++i)
    #define CLR(A,value) memset(A,value,sizeof(A))
    #define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
    
    
    //OTHER
    #define SZ(V) (int)V.size()
    #define PB push_back
    #define MP make_pair
    #define all(x) (x).begin(),(x).end()
    
    
    //INPUT
    #define RI(n) scanf("%d", &n)
    #define RII(n, m) scanf("%d%d", &n, &m)
    #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
    #define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
    #define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
    #define RS(s) scanf("%s", s)
    
    
    //OUTPUT
    #define WI(n) printf("%d\n", n)
    #define WS(n) printf("%s\n", n)
    
    
    //debug
    //#define online_judge
    #ifndef online_judge
    #define debugt(a) cout << (#a) << "=" << a << " ";
    #define debugI(a) debugt(a) cout << endl
    #define debugII(a, b) debugt(a) debugt(b) cout << endl
    #define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl
    #define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl
    #else
    #define debugI(v)
    #define debugII(a, b)
    #define debugIII(a, b, c)
    #define debugIV(a, b, c, d)
    #endif
    
    #define sqr(x) (x) * (x)
    typedef long long LL;
    typedef vector <int> VI;
    const int INF = 0x3f3f3f3f;
    const double EPS = 1e-10;
    const int MOD = 100000007;
    const int MAXN = 100010;
    const double PI = acos(-1.0);
    
    int prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 57};
    
    LL ipt, dfs_n, dfs_ans;
    int maxfac;
    
    void dfs(int index, int cnt, int pre, int ans, LL mul)
    {
        if (cnt == maxfac || mul > ipt / prime[index])
        {
            if (ans > dfs_ans)
            {
                dfs_ans = ans;
                dfs_n = mul;
            }
            if (ans == dfs_ans && mul < dfs_n)
            {
                dfs_n = mul;
            }
            return;
        }
        LL t = prime[index];
        FE(i, 1, min(pre, maxfac - cnt))
        {
            if (mul > ipt / t) return;
            dfs(index + 1, cnt + i, i, ans * (i + 1), mul * t);
            if (t > ipt / prime[index]) break;
            t *= prime[index];
        }
    }
    
    int main()
    {
        int kase;
        RI(kase);
        while (kase--)
        {
            dfs_ans = -1;
            scanf("%I64d", &ipt);
            maxfac = (int)log2(ipt * 1.0);
            dfs(0, 0, INF, 1, 1);
            printf("%I64d %I64d\n", dfs_n, dfs_ans);
        }
        return 0;
    }

[Link 1]: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=35214#problem/G