查找两个数字的LCM
在数学中，最小公倍数(LCM)是可能的最小整数，可以被两个数整除。
LCM可以通过许多方法来计算，例如因式分解等。但是在此算法中，我们将较大的数字乘以1，2，3…。直到找到一个可被第二个数字整除的数字。
输入输出
Input: Two numbers: 6 and 9 Output: The LCM is: 18
算法
LCMofTwo(a, b)
输入：两个数字a和b，视为a>b。
输出： a和b的LCM。
Begin
lcm := a
i := 2
while lcm mod b ≠ 0, do
lcm := a * i
i := i + 1
done
return lcm
End示例
#include<iostream>
using namespace std;
int findLCM(int a, int b) { //assume a is greater than b
int lcm = a, i = 2;
while(lcm % b != 0) { //try to find number which is multiple of b
lcm = a*i;
i++;
}
return lcm; //the lcm of a and b
}
int lcmOfTwo(int a, int b) {
int lcm;
if(a>b) //to send as first argument is greater than second
lcm = findLCM(a,b);
else
lcm = findLCM(b,a);
return lcm;
}
int main() {
int a, b;
cout << "Enter Two numbers to find LCM: "; cin >> a >> b;
cout << "The LCM is: " << lcmOfTwo(a,b);
}输出结果
Enter Two numbers to find LCM: 6 9 The LCM is: 18