Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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.
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.
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
3 5 7 15
6 4 10296 936 1287 792 1
Sample Output
105
10296
这道题就是要求多个数的最小公倍数；
思路很简单，就是每两个数求他们的最小公倍数，然后用求出的最小公倍数与下一个数
#include<stdio.h>
int gcd(int a,int b)
{
if(b==) return a;
return gcd(b,a%b);
}
int main()
{
int t;
int n,a,b,i;
int cnt;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
cnt=a=;
for(i=;i<=n;i++)
{
scanf("%d",&b);
cnt=a/gcd(a,b)*b;
a=cnt;
}
printf("%d\n",cnt);
}
return ;
}