【AOJ-41】Least Common Multiple

时间：2014-01-14 19:42:50 收藏：0 阅读：541

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 file 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

Original

Transformed

2
3 5 7 15
6 4 10296 936 1287 792 1

Sample Output

105
10296

思路：

每两个两个求最小公倍数

参考代码：

#include <stdio.h>
int 
GCD(int 
a,int 
b)//最大公约数
{
    if(!b)
        return 
a;
    return 
GCD(b,a%b);
}
 
int 
LCM(int 
a,int 
b)//最小公倍数
{
    return 
a/GCD(a,b)*b;
}
 
int 
a[10000];
 
int 
main()
{
    int 
t;
    scanf("%d",&t);
    while(t--)
    {
        int 
n;
        scanf("%d",&n);
        int 
i;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        int 
temp=a[0];
        for(i=1;i<n;i++)
            temp=LCM(temp,a[i]);
        printf("%d\n",temp);
    }
    return 
0;
}

原文：http://www.cnblogs.com/ahu-shu/p/3512717.html