Problem 5
问题描述：
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?


求出1~20的最小公倍数.

思路如下：

//辗转相除法求最大公倍数public long count_gcd(long a, long b){long big =  a>b ? a : b;long small = a>b ? b : a;long gcd = small;long r1, r2;while(big%small!=0){gcd = big%small;big = small;small = gcd;}return gcd;}public long solve1(int number){long result = 1;long gcd = 1;long lcm = 1;long a = 2;long b;for(int i=3; i<=number;i++){b = i;gcd = count_gcd(a,b);lcm = (a*b)/gcd;a = lcm;}return lcm;}