数论常用内容——反素数

提到反素数,大家可能比较陌生,这里博主我对反素数的了解也不是很深刻,希望能借此机会来总结一下并和大家交流

概念

对于任何正整数n,其约数个数为f(n),如果某个正整数n满足:对任意正整数i(0

性质

性质1、一个反素数的所有质因子必然是从2开始的连续若干个质数,因为反素数是保证约数个数为x的这个数n尽量小

性质2、如果n=2^t1*3^t2*5^t3*…,那么必有t1>t2>t3…

应用

1.给出一个数n,求一个最小的正整数x,使得x的约数个数为n

2.求出1~n中约数最多的这个数(构建搜索树,每层为素因子的i次幂,其中0<=i<=63,一般取63,因为2^64为long long上限,每层碰到所剩数字/该层代表素因子小于上层质因子时不再拓展本枝)

反素数表

int a[42] = {1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,554400,665280,720720,1081080,1441440,2162160};

int b[42] = {1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200,216,224,240,256,288,320};

应用示例

求一个最小正整数,使得它的因子个数为n

typedef unsigned long long ULL;
const ULL INF = ~0ULL;

int p[16] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};

int n;
ULL ans;

void dfs(int dept,ULL tmp,int num){
    if(num > n) return;
    if(num == n && ans > tmp) ans = tmp;
    for(int i=1;i<=63;i++){
        if(ans / p[dept] < tmp) break;
        dfs(dept+1,tmp *= p[dept],num*(i+1));
    }
}

int main(){
    while(cin>>n){
        ans = INF;
        dfs(0,1,1);
        cout<<ans<<endl;
    }
    return 0;
}

求n以内的因子数最多的数

typedef unsigned long long ULL;
const ULL INF = ~0ULL;

int p[16] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};

ULL ans,n;
int best;

void dfs(int dept,ULL tmp,int num){
    //到叶子结点,返回
    if(dept >= 16) return;
    //num记录的因子个数,如果遇到更小的,就更新
    if(num > best){
        best = num;
        ans = tmp;
    }
    //当因子个数相同时,取值最小的
    if(num == best && ans > tmp) ans = tmp;
    for(int i=1;i<=63;i++){
        if(n / p[dept] < tmp) break;
        dfs(dept+1,tmp *= p[dept],num*(i+1));
    }
}

int main(){
    while(cin>>n){
        ans = INF;
        best = 0;
        dfs(0,1,1);
        cout<<ans<<endl;
    }
    return 0;
}

超大数据(10^80)找n以内的约数最多的数并输出因子个数(java实现)

package main;

import java.io.*; 
import java.math.*; 
import java.util.*; 

class Node { 
    private static final int MAXP = 60; 

    public BigInteger K; 
    public long F; 
    public int N; 
    public int[] A; 

    public Node() { 
        K = BigInteger.ZERO; 
        A = new int[MAXP]; 
    } 
} 

public class Main{ 
    private static final int MAXIP = 250; 
    private static final int MAXP = 60; 

    private static BigInteger[] prime; 

    private static void init() { 
        boolean[] isPrime = new boolean[MAXIP]; 
        for(int i=0;i<MAXIP;++i){ 
            isPrime[i] = true; 
        } 
        isPrime[0] = isPrime[1] = false; 
        for(int i=4;i<MAXIP;i+=2) { 
            isPrime[i] = false; 
        } 
        for(int i=3;i<MAXIP;i+=2) { 
            for(int j=3;i*j<MAXIP;j+=2) { 
                isPrime[i*j] = false; 
            } 
        } 
        prime = new BigInteger[MAXP]; 
        for(int i=0, j=0;i<MAXIP;++i) { 
            if(isPrime[i]) { 
                prime[j++] = BigInteger.valueOf(i); 
            } 
        } 
    } 

    public static void main(String args[]) { 
        init(); 
        List<BigInteger> P = new ArrayList<BigInteger>(); 
        BigInteger MP = BigInteger.ZERO; 
        List<Node> ans = new ArrayList<Node>(); 
        Scanner cin = new Scanner(new BufferedInputStream(System.in)); 
        //Integer cas = new Integer(3);
        while(cin.hasNext()) //cas>0{ 
            BigInteger temp = cin.nextBigInteger(); 
            P.add(temp); 
            if(temp.compareTo(MP) == 1) { 
                MP = temp; 
            } 
            ans.add(new Node()); 
            //cas--;
        } 
        Map<Long, BigInteger> map = new HashMap<Long, BigInteger>(); 
        Queue<Node> queue = new LinkedList<Node>(); 
        Node origin = new Node(); 
        origin.K = BigInteger.ONE; 
        origin.F = 1; 
        origin.N = 0; 
        queue.add(origin); 
        map.put(origin.F, origin.K); 
        while(!queue.isEmpty()) { 
            Node u = queue.peek(); 
            queue.remove(); 
            BigInteger compare = map.get(u.F); 
            if(compare != null) { 
                if(compare.compareTo(u.K) == -1) { 
                    continue; 
                } 
            } 
            for(int i=0;i<P.size();++i) { 
                if(u.K.compareTo(P.get(i)) <= 0) { 
                    if(u.F > ans.get(i).F) { 
                        ans.get(i).F = u.F; 
                        ans.get(i).K = u.K; 
                    } 
                    else if(u.F == ans.get(i).F) { 
                        if(u.K.compareTo(ans.get(i).K) == -1) { 
                            ans.get(i).K = u.K; 
                        } 
                    } 
                } 
            } 
            for(int i=0;i<u.N;++i) { 
                Node v = new Node(); 
                v.K = u.K.multiply(prime[i]); 
                if(v.K.compareTo(MP) <= 0) { 
                    v.F = u.F / (u.A[i] + 1) * (u.A[i] + 2); 
                    v.N = u.N; 
                    for(int j=0;j<u.N;++j) { 
                        v.A[j] = u.A[j]; 
                    } 
                    ++ v.A[i]; 
                    boolean flag = true; 
                    compare = map.get(v.F); 
                    if(compare != null) { 
                        if(compare.compareTo(v.K) <= 0) { 
                            flag = false; 
                        } 
                        else { 
                            map.remove(v.F); 
                        } 
                    } 
                    if(flag) { 
                        queue.add(v); 
                        map.put(v.F, v.K); 
                    } 
                } 
            } 
            Node v = new Node(); 
            v.K = u.K.multiply(prime[u.N]); 
            if(v.K.compareTo(MP) <= 0) { 
                v.F = u.F * 2; 
                v.N = u.N + 1; 
                for(int i=0;i<u.N;++i) { 
                    v.A[i] = u.A[i]; 
                } 
                ++ v.A[u.N]; 
                boolean flag = true; 
                compare = map.get(v.F); 
                if(compare != null) { 
                    if(compare.compareTo(v.K) <= 0) { 
                        flag = false; 
                    } 
                    else { 
                        map.remove(v.F); 
                    } 
                } 
                if(flag) { 
                    queue.add(v); 
                    map.put(v.F, v.K); 
                } 
            } 
        } 
        for(int i=0;i<ans.size();++i) { 
            System.out.println(ans.get(i).K.toString() + " " + ans.get(i).F);
        } 
    } 
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值