ProjectEuler 005题

题目：

2520 is the smallest number that can be divided by each of the numbers from 1
to 10 without any remainder.

What is the smallest positive number that is evenly
divisible by all of the numbers from 1 to 20?

此题就是求最小公倍数的，刚开始我考虑的太复杂，打算求出每个数的素数因子，然后去处一些共有的部分，后来突然想到了最小公倍数。

 1 #include<iostream>

 2 using namespace std;

 3 int getLeastCommonMultiple(int num1, int num2);

 4 int GetMaxCommonDivide(int num1, int num2);

 5 int main()

 6 {

 7     int res = 20;

 8     for(int i = 19; i >=2; i--) {

 9         if( res % i == 0){

10             continue;

11         }

12         else {

13             res = getLeastCommonMultiple(res, i);

14         }

15     }

16     cout << res << endl;

17     system("pause");

18     return 0;

19 }

20 //求最小公倍数

21 int getLeastCommonMultiple(int num1, int num2) {

22     return num1*num2/(GetMaxCommonDivide( num1, num2));

23 }

24 /*  辗转相除法求最大公约数 */

25 int GetMaxCommonDivide(int num1, int num2) {

26     int temp;

27     while(num2!=0){

28         temp = num1%num2;

29         num1 = num2;

30         num2 = temp;

31     }

32     return num1;

33 }