题目：
Description
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)
结论是很简单的，不超过n的素数都会输出，超过n的都不会输出。
而且，2一定在里面，也就是说，2一定是第一个，这样就很方便了。
代码：
#include<iostream> #include<stdio.h> using namespace std; int degree_in_fact(int m, int p) { if (m)return degree_in_fact(m / p, p) + m / p; return 0; } bool isprime(int n) { for (int i = 3; i*i <= n; i += 2)if (n%i == 0)return false; return true; } int main() { int t, n; cin >> t; for (int cas = 1; cas <= t; cas++) { cin >> n; cout << "Case " << cas << ": " << n << " = 2 (" << degree_in_fact(n, 2) << ")"; for (int i = 3; i <= n; i += 2)if (isprime(i))
cout << " * " << i << " (" << degree_in_fact(n, i) << ")"; cout << endl; } return 0; }
不知道degree_in_fact这个函数的，请点击打开我的博客
LightOJ 1035 Intelligent Factorial Factorization
原文：http://blog.csdn.net/nameofcsdn/article/details/52247500