hdu 1019 least common multiple

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

 

思路：每两个数字求一次最大公约数，求最大公倍数的时侯要注意，见代码

<span style="font-family:Microsoft YaHei;">#include<stdio.h>int gcd(int a,int b){ int r; for(r=a%b;r!=0;r=a%b) {  a=b;  b=r; } return b;}int num[10100];int main(){ int n,m,i,j; int sum; while(scanf("%d",&n)!=EOF) {    while(n--)    {    scanf("%d",&m);    for(i=0;i<m;i++)    scanf("%d",&num[i]);     for(i=1;i<=m-1;i++)     {   sum=gcd(num[i],num[i-1]);   num[i]=(num[i]/sum)*num[i-1];//此处不能用这种方法num[i]=(num[i]*num[i-1])/sum，会超出int型的数据范围，     }     printf("%d\n",num[m-1]);    } } return 0;}</span>