Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41182 Accepted Submission(s): 15514
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 1010
#define ll long long
using namespace std;
int a[N];
ll gcd(ll x,ll y)
{
ll t;
if(x<y){t=x;x=y;y=t;}
return y?gcd(y,x%y):x;
}
int main()
{
int t,n,i,j;
ll sum;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
sum=1;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum=sum*a[i]/gcd(sum,a[i]);
}
printf("%lld\n",sum);
}
return 0;
}
Lowest Common Multiple Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45260 Accepted Submission(s): 18664
Problem Description
求n个数的最小公倍数。
Input
输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。
Output
为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。
Sample Input
2 4 6
3 2 5 7
Sample Output
12
70
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 1010
#define ll long long
using namespace std;
int a[N];
ll gcd(ll x,ll y)
{
ll t;
if(x<y){t=x;x=y;y=t;}
return y?gcd(y,x%y):x;
}
int main()
{
int t,n,i,j;
ll sum;
while(scanf("%d",&n)!=EOF)
{
sum=1;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum=sum*a[i]/gcd(sum,a[i]);
}
printf("%lld\n",sum);
}
return 0;
}
