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AcWing 869. 试除法求约数

题目链接

https://www.acwing.com/activity/content/problem/content/938/

思路

基于算术基本定理，从最小质数开始往上除，直到不含这个因数，而和数都是由比它小的质数相乘得来，所以所有约束都是质数。

CODE

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

int n;

vector<int> get_divided(int n){
    vector<int> res;
    
    for(int i = 1; i <= n / i; ++i){
        if(n % i == 0){
            res.push_back(i);
        
            if(i != n / i) res.push_back(n / i);
        }
    }

    sort(res.begin(), res.end());

    return res;
}

int main()
{
    cin >> n;
    
    while (n -- ){
        int a;
        scanf("%d", &a);
        
        auto res = get_divided(a);
        
        for(auto t : res) cout << t << ' ';
        puts("");
    }
}

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AcWing 870. 约数个数

题目链接

https://www.acwing.com/activity/content/problem/content/939/

思路

还是基于算术基本定理，一个数可以分解为唯一因式：$$K=p{1}^{a1}+p{2}^{a2}+...+p{i}^{ai}$$
那么约数的个数就是公式：$$(a1+1)(a2+1)...(ai+1)$$

CODE

#include<vector>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;

const int mod = 1e9 + 7;
int n;

int main()
{
    cin >> n;
    
    unordered_map<int, int> primes;
    
    while (n -- ){
        int a;
        scanf("%d", &a);
    
        for(int i = 2; i <= a / i; ++i){
            while(a % i == 0){
                a /= i;
                primes[i]++;
            }
        }
            if(a > 1) primes[a]++;
    }
    
    ll res = 1;
    for(auto t : primes) res = res * (t.second + 1) % mod;
    
    cout << res << endl;
}

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AcWing 871. 约数之和

题目链接

https://www.acwing.com/activity/content/problem/content/940/

思路

还是基于算术基本定理：$$K=p{1}^{a1}+p{2}^{a2}+...+p{i}^{ai}$$由这个因式我们可以得到以下公式：$$N=(p{1}^0+p{1}^1+...+p{1}^{a1})(p{2}^0+p{2}^1+...p{2}^{a2})...(p{i}^0+p{i}^1+...+p{i}^{ai})$$这个公式展开就是每项约数的和。

详细推导可以参考以下视频：约数个数与约数之和

CODE

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;

const int mod = 1e9 + 7;
int n;

int main()
{
    cin >> n;
    
    unordered_map<int, int> primes;
    
    while (n -- ){
        int a;
        scanf("%d", &a);
        
        for(int i = 2; i <= a / i; ++i){
            while(a % i == 0){
                a /= i;
                primes[i]++;
            }
        }
        
        if(a > 1) primes[a]++;
    }
    
    ll ans = 1;
    for(auto p : primes){
        ll a = p.first, b = p.second;
        ll res = 1;
        while(b--) res = (res * a + 1) % mod;
        ans = ans * res % mod;
    }
    
    cout << ans << endl;
}

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AcWing 872. 最大公约数

题目链接

https://www.acwing.com/activity/content/problem/content/941/

思路

辗转相除法
具体证明请看 VCR

CODE

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int gcd(int a, int b){
    return b ? gcd(b, a % b) : a;
}

int main()
{
    int n;
    cin >> n;
    
    while(n--){
        int a, b;
        scanf("%d%d", &a, &b);
        
        cout << gcd(a, b) << endl;
    }
}