反素数学习笔记

反素数就是1-x中因子数最多的数中最小的数

首先前15个素数的累和就会爆1e18，然后2的60也会爆1e18

反素数一定由不超过15个素数，指数非严格递减生成，这样就有了例题1的搜索框架

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例题1：给一个数n，求最小的数x让x的因子数为n

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
    ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
void ex(){puts("No");exit(0);}
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3fLL;
const ll Mod=1e18+7;
const double eps=1e-6;
const double pi=acos(-1.0);


const int prime[20]={1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};

int n;
ll ans;

void dfs(int dep,ll num,int cnt){
    if(cnt>n)return;
    if(cnt==n && ans>num)ans=num;
    for(int i=1;i<=60;i++){
        if(ans/prime[dep] < num)break;
        
        num*=prime[dep];

        dfs(dep+1,num,cnt*(i+1));
    }
}

int main(){
    scanf("%d",&n);
    ans=9e18;

    dfs(1,1,1);
    printf("%lld
",ans);
}

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例题2：多组的例题1(x)，加上了指数降级约束的版本

//#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //

#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
    ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
void ex(){puts("No");exit(0);}
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3fLL;
const ll Mod=1e18+7;
const double eps=1e-6;
const double pi=acos(-1.0);


const int prime[20]={1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};

ll n;
ll ans;
int bst_cnt;

void dfs(int dep,int limit,ll now,int cnt){
    if(now>n)return;
    if(cnt>bst_cnt){
        bst_cnt=cnt;
        ans=now;
    }
    if(cnt==bst_cnt && ans>now){ans=now;}
    for(int i=1;i<=limit;i++){
        double cur=(double)now;

        if(n<cur*prime[dep])break;
        dfs(dep+1,i,now*=prime[dep],cnt*(i+1));
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%lld",&n);
        ans=Linf;
        bst_cnt=0;
        dfs(1,60,1,1);
        printf("%lld %d
",ans,bst_cnt);
    }    
}