[Project Euler] Problem 5

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?

题目的要求就是求1，2，3 ··· ··· 20的最小公倍数

我们可惜先求1，2的最小公倍数2

再求2，3的最小公倍数6

再求6，4的最小公倍数12

... ...

而求最小公倍数，我们可以用两数的积除以它们的最大公约数

最大公约数我们用辗转相除法

#include <iostream>
using namespace std;

int getLCM(int,int);
int getGCD(int,int);

int main(){
    int multiple = 1;
    for(int i=1; i<=20; i++){
        multiple = getLCM(multiple,i);
    }
    cout << multiple << endl;
    return 0;    
}

int getLCM(int a,int b){
    int c = getGCD(a,b);
    return a/c*b;
}

int getGCD(int a,int b){
    int tmp;
    while(a%b != 0){
        tmp = a;
        a = b;
        b = tmp%b;
    }
    return b;
}