wenbao与反素数

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·

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打表

#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 500010;
const int M = 1024;

struct node {
    int a, b;
    node() {}
    node(int x, int y) : a(x), b(y) {}
    bool operator < (const node cmp) const {
        return b < cmp.b;
    }
}f[M];

bool vis[M];
int prime[M];
int cnt, num;

void get_prime() {
    int i, j;
    memset(vis, true, sizeof(vis));
    for(i = 2; i < M; ++i) {
        for(j = i*i; j < M; j += i) {
            vis[j] = false;
        }
    }
    cnt = 0;
    for(i = 2; i < M; ++i) {
        if(vis[i])  prime[cnt++] = i;
    }
}

void solve(int pnum, int val, int pos, int lim) {
    if(val > N) return ;
    f[num++] = node(pnum, val);
    LL nlim, nval, npnum;
    nval = val; nlim = 0; npnum = 1;
    while(nlim < lim && nval <= N) {
        nlim++; npnum++; nval *= prime[pos];
        if(nval <= N)   solve(npnum*pnum, nval, pos + 1, nlim);
    }
}
void get_antiprime() {
    get_prime();
    num = 0;
    solve(1, 1, 0, 32);
    sort(f, f + num);
    for(int i = 0; i < num; ++i){
        printf("%d--%d
", f[i].a, f[i].b);
    }
}
int main(){
    get_antiprime();
    return 0;
}


const int antiprime[]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,  
                       1260,1680,2520,5040,7560,10080,15120,20160,25200,  
                       27720,45360,50400,55440,83160,110880,166320,221760,  
                       277200,332640,498960,554400,665280  
                      };  
  
const int factorNum[]={1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,  
                       64,72,80,84,90,96,100,108,120,128,144,160,168,180,  
                       192,200,216,224  
                      }; 

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27
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29
30
31
#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 0x7fffffff
#define ll long long 
using namespace std;
int n,ans=1,num=1;
int p[15]={1,2,3,5,7,11,13,17,19,23,29,31};
void dfs(int k,ll now,int cnt,int last)
{
    if(k==12)
    {
        if(now>ans&&cnt>num){ans=now;num=cnt;}
        if(now<=ans&&cnt>=num){ans=now;num=cnt;}
        return;
    }
    int t=1;
    for(int i=0;i<=last;i++)
    {
        dfs(k+1,now*t,cnt*(i+1),i);
        t*=p[k];
        if(now*t>n)break;
    }
}
int main()
{
    scanf("%d",&n);
    dfs(1,1,1,20);
    printf("%d",ans);
    return 0;
}

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