acwing算法基础之数学知识--求一个数x的约数数目和约数之和

1 基础知识

求一个数x的约数数目和约数之和的关键步骤：

1. 对数x分解质约数，
   $$x=p_1^{c_1}\cdot p_2^{c_2}\cdots p_k^{c_k}$$

unordered_map<int,int> get_prime_divisors(int x) {//对一个数x进行分解质因子操作
	unordered_map<int,int> mp;
	for (int i = 2; i <= x / i; ++i) {
		if (x % i == 0) {
			int s = 0;
			while (x % i == 0) {
				x /= i;
				s++;
			}
			mp[i] = s;
		}
	}
	if (x > 1) mp[x] = 1;
	return mp;
}

2. 那么约数总数计算如下，
   $$cnt=(c_1+1)(c_2+1)\cdots (c_k+1)$$
3. 那么约数之和计算如下，
   $$s=(p_1^0+p_1^1+\cdots +p_1^{c_1})(p_2^0+p_2^1+\cdots +p_2^{c_2})\cdots (p_k^0+p_k^1+\cdots +p_k^{c_k})$$

2 模板

如果 N = p1^c1 * p2^c2 * ... *pk^ck
约数个数： (c1 + 1) * (c2 + 1) * ... * (ck + 1)
约数之和： (p1^0 + p1^1 + ... + p1^c1) * ... * (pk^0 + pk^1 + ... + pk^ck)

3 工程化

题目1：求一串数的乘积的约数总数。

#include 
#include 
#include 

using namespace std;

const int mod = 1e9 + 7;
unordered_map<int,int> mp;

void get_prime_divisors(int x) {
    //cout << "x = " << x << endl;
    for (int i = 2; i <= x / i; ++i) {
        if (x % i == 0) {
            int s = 0;
            while (x % i == 0) {
                x /= i;
                s++;
            }
            mp[i] += s;
            //cout << "i = " << i << ", s = " << s << endl;
        }
    }
    if (x > 1) {
        mp[x] += 1;
        //cout << "x = " << x << ", mp[x] = " << mp[x] << endl;
    }
    return;
}

int main() {
    int n;
    cin >> n;
    while (n--) {
        int x;
        cin >> x;
        get_prime_divisors(x);
    }
    
    long long res = 1;
    for (auto [x, y] : mp) {
        res *= (y + 1);
        res %= mod;
    }
    cout << res << endl;
    
    return 0;
}

题目2：求一串数的约数之和。

#include 
#include 

using namespace std;

const int mod = 1e9 + 7;
unordered_map<int,int> mp;

void get_prime_divisors(int x) {
    for (int i = 2; i <= x / i; ++i) {
        if (x % i == 0) {
            int s = 0;
            while (x % i == 0) {
                x /= i;
                s++;
            }
            mp[i] += s;
        }
    }
    if (x > 1) mp[x] += 1;
    return;
}

int main() {
    int n;
    cin >> n;
    while (n--) {
        int x;
        cin >> x;
        
        get_prime_divisors(x);
    }
    
    long long res = 1;
    for (auto [x, a] : mp) {
        //遍历每一个质数x
        //cout << "x = " << x << ", a = " << a << endl;
        long long t = 1;
        while (a--) {
            t = (t * x + 1) % mod;
        }
        res = (res * t) % mod;
    }
    cout << res << endl;
    
    return 0;
}