技術標籤：數學演算法

本文介紹約數個數定理，一個數的約數是可以計算的。

···
參照百度百科-約數個數定理

假設：正整數378000共有多少個正約數？
解：將378000分解質因數378000=2⁴×3³×5³×7¹
由約數個數定理可知378000共有正約數(4+1)×(3+1)×(3+1)×(1+1)=160個。

···
練習題-Problem
簡單的思路就是用雜湊表存下<底數， 指數>
最後再將指數求積即可，注意每次求積都要求餘，否則會移除（即使long也如此）

Accepted Code:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.security.KeyStore.Entry;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main
 
 {

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(System.in), 1 << 15);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null ||
 
 !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}
	}

	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	//  Input class by SoKee learned from WC:Petr and thank you
	///
	static final int INF = 0x3F3F3F3F;
	static final int MOD_1e9plus7 = (int) (1e9 + 7);
	static HashMap<Integer, Integer> map = new HashMap<>();
	static int n, x;
	static long res = 1;

	public static void main(String[] args) {

		n = in.nextInt();
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			bigAdd(x);
		}
		
		work();
		out.println(res % MOD);
		out.flush();
	}

	static void bigAdd(int x) {
		for (int i = 2; i <= x / i; i++) {
			if (x % i == 0) {
				int s = 0;
				while (x % i == 0) {
					x /= i;
					s++;
				}
				map.put(i, map.getOrDefault(i, 0) + s);
			}
		}
		if (x > 1)
			map.put(x, map.getOrDefault(x, 0) + 1);
	}

	static void work() {
		for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
			res = res * (entry.getValue() + 1) % MOD;
		}
		return;
	}

}