Divisors（约数的数目）

UVA294

题目要求指定范围内约数数目最多的数以及对应的约数数量。
根据唯一分解定理，对于任意正整数 $2\leq N$，有：
$N=\sum\limits_{i=1}^{n}p_i^{a_i}，p为素数$
那么根据乘法原理，约数的总数目为：
$f(N)=\prod\limits_{i=1}^{n}(1+a^i)$

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

const int MAXN = 1000001;

int main() {
	int T;
	cin >> T;
	int Left, Right;
	while (T--) {
		cin >> Left >> Right;
		int Ans = 0, Node = Left;
		for (int n = Left; n <= Right; ++n) {
			int Number = n;
			int Count = 1;
			int Max = ceil(sqrt(Right));
			int NumberOfDifP = 0;
			for (int i = 2; i <= Max; ++i) {
				if (Number % i == 0) {
					++NumberOfDifP;
					int temp = 0;
					while (Number % i == 0) {
						Number /= i;
						++temp;
					}
					Count *= (temp + 1);
				}
			}
			if (NumberOfDifP == 0) {
				Count = 1;
			}
			if (Number != 1) {
				++Count;
			}
			if (Ans < Count) {
				Ans = Count;
				Node = n;
			}
		}
		printf("Between %d and %d, %d has a maximum of %d divisors.\n", Left, Right, Node, Ans);
	}

	return 0;

}