Intelligent Factorial Factorization LightOJ

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

3

2

3

6

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 0x3F3F3F3Fusing 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;}