ACM hdu 1019 Least Common Multiple

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

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简单数学题，求连续n个数的最小公倍数

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Code:

#include <iostream>#include<string.h>using namespace std;int a[10000];int gcd(int a,int b)//求最大公约数{    int temp;    if(a<b)    {        temp = a;a = b;b = temp;    }    return (b==0)?a:gcd(b,a%b);}int LCM(int a,int b)//求最小公倍数{    return a/gcd(a,b)*b; //开始写成 a*b/gcd(a,b),WA了两次，因为可能会溢出}int main(){    int T,n,ans,x;    cin>>T;    while(T--)    {        //memset(a,0,sizeof(a));        ans = 1;        cin>>n>>x;        ans = LCM(ans,x);        for(int i = 0;i<n-1;i++)        {            cin>>x;            ans = LCM(ans,x);        }        cout<<ans<<endl;    }    return 0;}