HDU 1019 Least Common Multiple（求最小公倍数）

                                                             Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44106    Accepted Submission(s): 16553

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

 

题解：求最小公倍数。。。。

两数的最小公倍数=两数相乘 再除以两数的最大公约数 

AC代码：

#include<cstdio>#include<iostream>using namespace std;int gcd(int a,int b){if(!b)return a;return gcd(b,a%b);}int main(){int T,n,b,c;scanf("%d",&T);while(T--){scanf("%d%d",&n,&b);for(int i=1;i<n;i++){scanf("%d",&c);b=b/gcd(b,c)*c; //两数的最小公倍数=两数相乘再除以两数的最大公约数 }printf("%d\n",b);}return 0;}