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