hdu 1019 Least Common Multiple

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Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54584    Accepted Submission(s): 20824


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

 

Sample Output
105 10296
 
只要掌握基本的求最大公约数 与 最小公倍数的方法 即可简单解决该问题~

//求最小公倍数算法:

//最小公倍数=两整数的乘积 ÷最大公约数

//求最大公约数算法:

其中之一  --> 辗转相除法:

//有两整数a和b;

//① a%b得余数c

//② 若c=0,则b即为两数的最大公约数

//③ 若c≠0,则a=b,b=c,再回去执行①

 

代码如下:

#include <iostream>
#define N 10010
using namespace std;

long long gcd(long long a,  long long b)
{
  if(a % b == 0)
  {
    return b;
  }
  else
  {
    return gcd(b,a % b);
  }
}

long long num[N];
int main()

{
  int n, a;
  long long c;
  scanf("%d", &n);
  while(n--)
  {
    c= 1;
    scanf("%d", &a);
    for(int i = 0;i < a; ++i)
    {
      scanf("%I64d", &num[i]);
    }
    for(int i = 0;i < a; ++i)
    {
      if(c % num[i] == 0)
      {
        continue;
      }
      else
      {
        c = c*num[i]/gcd(c,num[i]);
      }
    }
    printf("%ld\n", c);
  }
  return 0;
}

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