HDOJ 1019 Least Common Multiple(最小公倍最大公约)

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 38281 Accepted Submission(s): 14413



[align=left]Problem Description[/align]
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

[align=left]Input[/align]
Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where
m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

[align=left]Output[/align]
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

[align=left]Sample Input[/align]

2
3 5 7 15
6 4 10296 936 1287 792 1

[align=left]Sample Output[/align]

105
10296

[align=left]Source[/align]
East Central North America 2003, Practice

[align=left]Recommend[/align]
JGShining | We have carefully selected several similar problems for you: 1005 1008 1061 1049 1108

#include<stdio.h>

int LCM(int a,int b)
{
if(b==0)return a;
else
return LCM(b,a%b);
}
int main(){
int n,m,i,a,b;
scanf("%d",&n);
while(n--){
scanf("%d%d",&m,&a);
for(i=1;i<m;i++)
{
scanf("%d",&b);
a=a/LCM(a,b)*b;//最小公倍数等于两个数的乘积除以最大公约数
}
printf("%d\n",a);
}
return 0;
}