HDU 1019 Least Common Multiple 最小公倍数 水题

Least Common Multiple

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

Total Submission(s): 48803    Accepted Submission(s): 18504


Problem Description

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.

 

Input

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.

 

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

 

Sample Input

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

 

Sample Output

105
10296

 

Source

East Central North America 2003, Practice

 

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题目要求就是求最小公倍数，还要求在32位内，我直接用的longlong 就可以过了
ac代码：

#include <stdio.h>

long long gcd(long long x,long long y)
{
int temp;

if(x<y)
{
temp=x;
x=y;
y=temp;
}
while(y!=0)
{
temp=x%y;
x=y;
y=temp;
}

return x;
}

int main()
{
long long num,n,ans,tmp,i;

scanf("%I64d",&num);
while(num--)
{
scanf("%I64d",&n);
ans=1;
for(i=0;i<n;i++)
{
scanf("%I64d",&tmp);
ans=(ans*tmp)/gcd(ans,tmp);
//	printf("***%I64d***\n",gcd(ans,tmp));
}
printf("%I64d\n",ans);
}

return 0;
}