标签:blog os io ar div amp log size
这题直接枚举是不可能的
所以想到了一遍输入一边计算 把每次的gcd给除掉 并相乘 得到的就是lcm
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
ll gcd(ll a,ll b)
{
return a%b==0?b:gcd(b,a%b);
}
int main()
{
int t,n,tn;
ll a,sum;
cin>>t;
while(t--)
{
cin>>n;
tn=n;
sum=1;
while(n--)
{
//cout<<n<<endl;
scanf("%I64d",&a);
if(n==tn-1) {sum=a;continue;}
sum=sum*a/(gcd(sum,a));
}
cout<<sum<<endl;
}
return 0;
}
hdu 1019 Least Common Multiple,布布扣,bubuko.com
hdu 1019 Least Common Multiple
标签:blog os io ar div amp log size
原文地址:http://www.cnblogs.com/sola1994/p/3923656.html