HDU 1019 Least Common Multiple

題目大意：

給你T組資料，開頭給你一個n，求後面n個數字的最小公倍數

解題思路：

這道是一道簡單的lcm板子題，直接暴力就水過了

程式碼如下：

 1 #include <bits/stdc++.h>
 2 #define mem(a) memset(a,0,sizeof(a))
 3 #define mem1(a) memset(a,-1,sizeof(a))
 4 #define forn(i,n) for(int i=0;i<n;++i)
 5 #define for1(i,n) for(int i=1;i<=n;++i)
 6
 
 #define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
 7 #define ll long long
 8 #define LL long long
 9 #define inf 0x3f3f3f3f
10 #define INF 0x3f3f3f3f3f3f3f3f
11 #define N 1000000
12 using namespace std;
13 const int maxn=1e3+50;
14 ll gcd(ll a, ll b){return b ? gcd(b, a % b) : a;}
15 ll lcm(ll a, ll b){return
 
 a * b / gcd(a, b);}
16 
17 int num[maxn];
18 
19 int main()
20 {
21     int T,n,m;
22     scanf("%d",&T);
23     while(T--)
24     {
25         ll tmp,ans=1;
26         scanf("%d",&n);
27         for1(i,n)
28         {
29             scanf("%lld",&tmp);
30             ans=lcm(ans,tmp);
31         }
 
32         printf("%lld\n",ans);
33     }
34     return 0;
35 }