求一個數的約數個數 d(n)

·方法一

·方法二

·時間測試

 

方法一：篩法

for(int i = 1; i <= n; ++i)
    for(int j = 1; i * j <= n; ++j)
        d[i * j]++;

方法二：質因數分解

若A|B 則

a1<=b1,a2<=b2···an<=bn (a,b分別指正數A,B的質因數)

則A可取 (0~a1)*(0~a2)*···*(0~an)

即 (a1+1)*(a2+1)*···*(an+1)

#include <cstdio>
#include <cstring>
#define N 100001

int n,p;
bool mark[N];
int prime[N];

void get_prime(){
    for(int i = 2; i <= N;++i){
        if(!mark[i]) prime[++p] = i;
        for(int j = 1; j <= p; ++j){
            if(i * prime[j] > N) break;
            mark[i * prime[j]] = true;
            if(i % prime[j] == 0) break;
        }
    }
}

int get(int n1){
    int w1[N];
    memset(w1, 0, sizeof(w1));
    for(int i = 1; prime[i] <= n1; ++i){
        while(n1 % prime[i] == 0){
            w1[i]++;
            n1 = n1 / prime[i];
        }
    }
    int ans = 1;
    for(int i = 1; i <= p; ++i) ans = ans * (w1[i] + 1);
    return ans;
}

int main(){
    get_prime();
    printf("%d", get(36));
    //for(int i = 1; i <= p; ++i) printf("%d ", prime[i]);
    return 0;
}

 時間測試：

#include <bits/stdc++.h>

using namespace std;

const int N = 100001;

int n,p;
int prime[N],mark[N],d[N];

void od(){
    for(int i = 1; i <= n; ++i){
        for(int j = 1; i * j <= n; ++j){
            d[i * j]++;
        }
    }
    /*for(int i = 1; i <= n; ++i){
        for(int j = 1; i * j <= n; ++j){
            d[i * j].push_back(i);//約數 
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 1; i * j <= n; ++j){
            d[i * j] += i;           //約數和 
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 2; i * j <= n; ++j){
            d[i * j] += i;           //約數和（不包自身） 
        }
    }*/
} 

void get_prime(){
    for(int i = 2; i <= n; ++i){
        if(!mark[i]) prime[++p] = i;
        for(int j = 1; j <= p; ++j){
            if(i * prime[j] > n)  break;
            mark[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
}

int od1(){
    int w1[N];
    int pp = 0, n1 = n;
    for(int i = 1; i <= p && prime[i] <= n1; ++i){
        if(n1 % prime[i] == 0) w1[++pp] = 0;
        while(n1 % prime[i] == 0){
            n1 /= prime[i];
            w1[pp]++;
        }
    }
    int ans = 1;
    for(int i = 1; i <= pp; ++i) ans *= w1[i] + 1;
    return ans;
}

int main(){
    clock_t start, end;
    n = 99999;
    cout << "n = " << n << endl;
    
    start = clock();
    od();
    end = clock();
    cout << "篩法時間:" << (double)(end - start) / CLOCKS_PER_SEC << "秒" << endl; 
    cout << "d(n) = " << d[n] << endl;
    
    start = clock();
    get_prime();
    int ans = od1();
    end = clock();
    cout << "質因數分解時間:" << (double)(end - start) / CLOCKS_PER_SEC << "秒" << endl;
    cout << "d(n) = " << ans << endl;
    
    return 0;
}