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 }
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原文：http://www.cnblogs.com/wanghui390/p/3749417.html