HDU 1019 Least Common Multiple 数学题

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

 

中文意思就是给你n组数据，让你输出其中的最小公倍数。

意思很简单，只要用辗转相除法求出两数的最大公因数，则两数的最小公倍数就是两数之积除以两数的最大公倍数即可。

不过不能直接用int，数据给的很大，要用long long。

代码如下：

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<cstring>
 5 using namespace std;
 6 int n,t;
 7 long long nn,ans;
 8 int gcd(long long x,long long y)
 9 {
10     long long a;
11      a=x>y? x:y;
12      while(y!=0)
13      {
14          a=x%y;
15          x=y;
16          y=a;
17      }
18      return x;
19 }
20 int main(){
21    scanf("%d",&n);    
22     for(int i=1;i<=n;++i)
23     {
24         scanf("%d",&t);
25         ans=1;
26         for (int j=1;j<=t;++j)
27         {
28             scanf("%I64d",&nn);
29             ans=(ans*nn)/gcd(ans,nn);
30             
31         }
32         printf("%d\n",ans);
33     }
34     return 0;
35 }