Factovisors - PC110704

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原文地址：http://www.milkcu.com/blog/archives/uva10139.html

原创：

作者：MilkCu

题目描述

Problem D: Factovisors

The factorial function, n! is defined thus for n a non-negative integer:

   0! = 1   n! = n * (n-1)!   (n > 0)

We say that a divides b if there exists an integer k such that

   k*a = b

The input to your program consists of several lines, each containing two non-negative integers, n and m, both less than 2^31. For each input line, output a line stating whether or not m divides n!, in the format shown below.

Sample Input

6 96 2720 1000020 1000001000 1009

Output for Sample Input

9 divides 6!27 does not divide 6!10000 divides 20!100000 does not divide 20!1009 does not divide 1000!

解题思路

思路比较流畅，依次求2~n之间的数与m的最大公约数，然后用公约数去除m。
若m能被除尽，则m可以整除n!；否则不能。
但是提交到UVaOJ为什么会超时呢？看来还需要优化。

如果在遍历2~n之间的数时，遇到的数i为质数。
若m为素数且m>n，则m与2~n中的任何数互素。
其他优化：把开方放在循环外边。
从n到2递减寻找最大公约数可以提高效率。

还要注意特殊情况的处理。

PC可以通过，UVaOJ总是超时，太浪费时间了。
对于我的不高的要求，以后使用PC吧。

两种求最大公约数的方法：

1. 递归

int gcd(int a, int b) {if(b == 0) {return a;}if(a < b) {return gcd(b, a);}return gcd(b, a % b);}

2. 循环

int gcd(int a, int b) {if(a < b) {int tmp = a;a = b;b = tmp;}while(b > 0) {int tmp = a;a = b;b = tmp % b;}return a;}

代码实现

#include <iostream>#include <algorithm>#include <cmath>using namespace std;int isPrime(int x) {int sq = sqrt(x);for(int i = 2; i <= sq; i++) {if(x % i == 0) {return 0;}}return 1;}int gcd(int a, int b) {if(a < b) {int tmp = a;a = b;b = tmp;}while(b > 0) {int tmp = a;a = b;b = tmp % b;}return a;}int main(void) {int n, m;while(cin >> n >> m) {if(m == 0) {cout << m << " does not divide " << n << "!" << endl;continue;}if(m == 1) {cout << m << " divides " << n << "!" << endl;continue;}if(isPrime(m) && m > n) {cout << m << " does not divide " << n << "!" << endl;continue;}int tm = m;for(int i = n; i >= 2; i--) {int g = gcd(i, tm);if(g > 1) {//cout << i << "  " << g << endl;tm /= g;}if(tm == 1) {cout << m << " divides " << n << "!" << endl;break;}}if(tm != 1) {cout << m << " does not divide " << n << "!" << endl;}}return 0;}

（全文完）

本文地址：http://blog.csdn.net/milkcu/article/details/23592449

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