HDU 1019 Least Common Multiple(最小公倍数&最大公约数熟悉)

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42699 Accepted Submission(s): 16044Problem DescriptionThe 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.InputInput 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 integersin the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.OutputFor 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

关于最小公倍数和最大公约数1、最大公约是

int GCM(int a,int b)
{
return b==0?a:GCM(b,a%b);
}

2、最小公倍数

任意两个数的最小公倍数：即用两个数的乘绩除以两个数的最大公约数

代码一：

#include <iostream>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;int GCM(int a,int b)
{
return b==0?a:GCM(b,a%b);
}int main(){int i,j,m,n,t,cnt,k;int num[1001];cin>>t;while(t--){cin>>n;for(i=0;i<n;i++)cin>>num[i];if(n>1){k = num[0]*(num[1]/GCM(num[0],num[1]));for(i=2;i<n;i++){k = k*(num[i]/GCM(k,num[i]));}}elsek = num[0];cout<<k<<endl;}}

代码二 暴力：

#include <iostream>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;int num[100001];int main(){int i,j,m,n,t,cnt,k;cin>>t;while(t--){cin>>n;for(i=0;i<n;i++)cin>>num[i];sort(num,num+n);for(i=1;;i++){for(j=n-2;j>=0;j--){if(num[n-1]*i % num[j] != 0)break;}if(j==-1){cout<<num[n-1]*i<<endl;break;}}}}