有哪位做过求最小公倍数和最大公约数的题目？如何在现有变量下完成两个数的求值呢？题目稍后上。

程序填空，不要改变与输入输出有关的语句。
输入一个正整数 repeat (0<repeat<10)，做 repeat 次下列运算：
输入两个正整数m和n，输出它们的最小公倍数和最大公约数。
输入输出示例：括号内为说明
输入：
3 (repeat=3)
3 7 (m=3,n=7)
24 4 (m=24,n=4)
24 18 (m=24,n=18)
输出：
21 is the least common multiple of 3 and 7, 1 is the greatest common divisor of 3 and 7.
24 is the least common multiple of 24 and 4, 4 is the greatest common divisor of 24 and 4.
72 is the least common multiple of 24 and 18, 6 is the greatest common divisor of 24 and 18.




#include <stdio.h>
int main(void)
{
int ***, lcm, m, n;
int repeat, ri;

scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%d", &m);
scanf("%d", &n);
if(m <= 0 &#124;&#124; n <= 0)
printf("m <= 0 or n <= 0");
else{
/*---------*/
printf("%d is the least common multiple of %d and %d, %d is the greatest common divisor of %d and %d.\n", lcm, m, n, ***, m, n);
}
}
}



　　　　　　

***。。。这是变成违禁字符了？

　　　　　　

违禁词喜感，ccp

　　　　　　

#include <stdio.h>
int main(void)
{
int ***, lcm, m, n;
int repeat, ri,i; scanf("%d", &repeat);
for(ri = 1; ri <= repeat; ri++){
scanf("%d", &m);
scanf("%d", &n);
if(m <= 0 &#124;&#124; n <= 0)
printf("m <= 0 or n <= 0");
else{
lcm=m*n;
for(i=1;i<=lcm;i++)
if(i%m==0&&i%n==0){
lcm=i;
break;
}
if(m<n)
***=m;
else
***=n;
for(i=***;i>=1;i++)
if(m%i==0&&n%i==0){
***=i;
break;
}
printf("%d is the least common multiple of %d and %d, %d is the greatest common divisor of %d and %d.\n", lcm, m, n, ***, m, n);
}
}

　　　　　　

违禁词喜感，ccp

　　　　　　

违禁词。。。哈哈~~~

　　　　　　

Our Party g*d




　　　　　　

原本没有i这个变量的...
我想说在不添加变量的前提下,能否做出?



　　　　　　

额。。。这个，我看看袄。。。好久前做的，我直接复制上来的，都不知道有木有错