Intelligent Factorial Factorization LightOJ - 1035

智能阶乘质因数分解

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本文介绍了一个算法问题——智能阶乘质因数分解，该问题要求输入一个整数N，输出N!的所有质因数及其对应的指数。文章提供了完整的C++实现代码，并详细解释了如何处理多个测试用例，通过递归方式找到每个数的质因数。

题目

Intelligent Factorial Factorization

LightOJ - 1035

Given an integer N, you have to prime factorize N! (factorial N).

Input

Input starts with an integer T (≤ 125), denoting the number of test cases.

Each case contains an integer N (2 ≤ N ≤ 100).

Output

For each case, print the case number and the factorization of the factorial in the following format as given in samples.

Case x: N = p₁ (power of p₁) * p₂ (power of p₂) * ...

Here x is the case number, p₁, p₂ ... are primes in ascending order.

Sample Input

Sample Output

Case 1: 2 = 2 (1)

Case 2: 3 = 2 (1) * 3 (1)

Case 3: 6 = 2 (4) * 3 (2) * 5 (1)

思路

素数分解。。。

代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <set>
#include <map>
#include <vector>

#define MAX_N 0x3F3F3F3F

using namespace std;

typedef long long ll;

map<int, int> pfactMap;
void primeFact(int n);

int main()
{
	int T, N;

	cin >> T;
	for (int i = 1; i <= T; i++) {
		cin >> N;
		pfactMap.clear();
		for (int n = N; n > 1; n--) {
			primeFact(n);
		}

		cout << "Case " << i << ": ";
		cout << N << " = ";
		map<int, int>::iterator it;
		int s;
		for (it = pfactMap.begin(), s = 0; it != pfactMap.end(); it++, s++) {
			cout << (it)->first << " (" << (it)->second << ")";
			if (s != pfactMap.size()-1)  cout << " * ";
		}
		cout << endl;
	}
    return 0;
}

void primeFact(int n) {
	for (int i = 2; i*i <= n; i++) {
		if (n % i == 0) {
//			pfactMap.insert(n%i);
			pfactMap[i]++;
			primeFact(n/i);
			return;
		}
	}
//	pfactMap.insert(n);
	pfactMap[n]++;
	return;
}