時間限制：1 秒

記憶體限制：128 兆

特殊判題：否

提交：6175

解決：1855

題目描述：

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.

輸入：

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.

輸出：

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

樣例輸入：

2
3 5 7 15
6 4 10296 936 1287 792 1

樣例輸出：

105

10296

本題求多個數的最小公倍數。GCD/LCM的問題都比較簡單，記住歐幾里得演算法就行了，LCM在算出GCD之後簡單處理一下即可。對於多個數，我是先把他們升序排序，然後兩兩求LCM，最後求得的一定是他們全部的LCM。需要注意的是數的範圍，本題要用long才不會出錯。

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
long gcd(long a,long b)
{
    if(b==0)    return a;
    return gcd(b,a%b);
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        int m;
        cin>>m;
        long *a=new long[m+1];
        memset(a,0,sizeof(a[0])*m);
        for(int i=0;i<m;i++)
            cin>>a[i];
        sort(a,a+m);
        long tmp=1;
        for(int i=0;i<m;i++)
            tmp=tmp*a[i]/gcd(tmp,a[i]);
        cout<<tmp<< endl;
    }
    return 0;
}
 

**********************************************

堅持，而不是打雞血~