hdu1019——Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42668    Accepted Submission(s): 16035

Problem Description

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.

 

Input

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.

 

Output

For 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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296

题目是求一些数的最小公倍数，用辗转相除法，主要是寒假复习，并不是什么难点！

#include<iostream>#include<cstring>#include<string>#include<queue>#include<cmath>#include<stack> #include<vector>#include<iomanip>#include<algorithm>#include<cctype>#include<cstdlib>#include<cstdio>using namespace std;int lcm(int a,int b){int r,x=a,y=b;while(b){r=a%b;a=b;b=r;}return x/a*y;}int main(){int n;cin>>n;while(n--){int tem=1,a,num[10000];cin>>a;for(int i=0;i<a;i++)cin>>num[i];for(int i=0;i<a;i++)tem=lcm(tem,num[i]);cout<<tem<<endl;}return 0;}

我只想做一个努力的人，加油！