http://acm.hdu.edu.cn/showproblem.php?pid=1108
最小公倍数
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32700 Accepted Submission(s): 18237
10 14
70
<span style="font-size:24px;">#include<iostream>
#include<cmath>
using namespace std;
int main()
{
long int a,b,m;
int i;
while(~scanf("%ld%ld",&a,&b))
{
if(a>b)
m=a;
else
m=b;
for(i=m;i<a*b;i+=m)//i+=m
{
if(i%a==0&&i%b==0)
{
break;
}
}
printf("%d\n",i);
}
return 0;
}</span>原文地址：http://blog.csdn.net/u012766950/article/details/38024271
