[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;
}