HDOJ 1019 Least Common Multiple (求n个数的LCM)

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42352    Accepted Submission(s): 15931

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
   
   
   
   

 
题意：求n个数最小公倍数

思路：题目不难，但是有几个小细节：
1.输入的数可能为0
2.更新ans(最小公倍数)的时候可能会爆int，所以说先除后乘，也可以用longlong


ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b)
{
	return b?gcd(b,a%b):a;
}
int main()
{
	int t,n,a,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		int ans;
		scanf("%d",&a);
		ans=a;
		int bz=0;
		for(i=1;i<n;i++)
		{
			scanf("%d",&a);
			if(bz)
			continue;
			if(a==0)
			ans=0,bz=1;
			int k=gcd(a,ans);
			if(k==0)
			ans=0;
			else
			ans=a/k*ans;
		}
		printf("%d\n",ans);
	}
	return 0;
}