一、循环求解

• First two LCM must be greater than the number of the larger number of,And less than the product of two Numbers
• 找出a与b之间的较大值max
• 利用循环,令i=max,不断++,当i能同时整除a与b时,输出i,跳出循环
• 即i为最小公倍数

代码如下：

#include <stdio.h>
int main() {
    
	int a, b;
	scanf("%d %d", &a, &b);
	int max = a > b ? a : b; //找出较大值
	int i = 0;
	for (i = max; i <= a*b; i++) {
    
		if (i%a == 0 && i%b == 0) {
    
			printf("%d\n", i);
			break;//跳出循环
		}
	}
	return 0;
}

二、辗转相除法

• 以除数和余数反复做除法运算,当余数为 0 时,取当前算式除数为最大公约数

• 两数乘积=最小公倍数*最大公约数

• By division algorithm to find the greatest common divisor that is the least common multiple

• 图解举例：

代码如下：

#include<stdio.h>
int main() {
    
	int a, b;
	scanf("%d %d", &a, &b);
	int m = a * b;
	int r = 0;
	while (r = a % b) {
    //r为0时,跳出循环,此时b为最大公因数
		a = b;
		b = r;
	}
	printf("%d\n", m/b);//Product divided by the greatest common factor is equal to the least common multiple
	return 0;
}

三、找最小 i 值

• There must be an integerk能够使k/a=i,k/b=j;
• a * iFor integerk为a的倍数,k%b为0表示k为b的倍数,即(a * i)%b==0,此时a和bCommon multiple ofk,即a * i;
• Minimum required LCM, find out to meet the requirements of i 值即可.

代码如下：

#include<stdio.h>
int main() {
    
	int a, b;
	scanf("%d %d", &a, &b);
	int i = 1;//i从1开始找
	while (a*i % b) {
    //(a*i)%b==0时为假,结束循环
		i++;
	}
	printf("%d\n", a*i);
	return 0;
}