UC知道好多不懂
来源：百度知道 编辑：UC知道 时间：2024/09/18 05:11:08
Least Common Multiple
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 89 Accepted Submission(s) : 22
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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
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 89 Accepted Submission(s) : 22
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
这不就求最小公倍数嘛
long int gcd(long int a,long int b)
{
long int t;
while(b)
{
t=a%b;
a=b;
b=t;
}
return a;
}
long int gcl(long a,long b)
{
return a*b/gcd(a,b);
}
void main()
{
long int n,m,a,t;
scanf("%ld",n);
while(n--)
{
scanf("%ld",&m);
scanf("%ld",&t);
for(i=0;i<m-1;i++)
{
scanf("%ld",&a);
t=gcl(a,t);
}
printf("%ld\n",t);
}
}
我也不懂