题目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：5958

解决：1774

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：

10510296

题目链接：http://ac.jobdu.com/problem.php?pid=1439

代码实现：

#include <iostream>#include<stdio.h>#include<string.h>/* run this program using the console pauser or add your own getch, system("pause") or input loop */using namespace std;int gcd(int a,int b){if( a== 0 && b==0){return -1;}else if(a ==0 || b==0){return a+b;}else {return gcd(b,a%b);}}int lcm(int a,int b){/*a、b的最小公倍数=（a*b）/a、b的最大公约数 */int gcdm=gcd(a,b);if(gcdm == -1){return -1;  //a=0 b=0 不存在最大公约数 }else{return a/gcdm*b;} }int main(int argc, char *argv[]) {int n;while(cin>>n){while(n--){int num[100];memset(num,0,100);int length;cin>>length;  //数字的个数for(int i=0;i<length;i++){cin>>num[i];} //3.两两求出LCMint ans;ans=num[0];for(int i=1;i<length;i++){ans=lcm(ans,num[i]);}cout<<ans<<endl;}}return 0;}