九度1439：Least Common Multiple

题目1439：Least Common Multiple

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：6175

解决：1855

题目描述：

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.

输入：

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.

输出：

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

样例输入：

23 5 7 156 4 10296 936 1287 792 1

样例输出：

105
10296 

本题求多个数的最小公倍数。GCD/LCM的问题都比较简单，记住欧几里得算法就行了，LCM在算出GCD之后简单处理一下即可。对于多个数，我是先把他们升序排序，然后两两求LCM，最后求得的一定是他们全部的LCM。需要注意的是数的范围，本题要用long才不会出错。

  

#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;}

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坚持，而不是打鸡血~