（HDOJ 1019）Least Common Multiple

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最新推荐文章于 2020-03-23 18:11:10 发布

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原文链接：http://www.cnblogs.com/cpoint/archive/2011/04/13/2015298.html

本文解析了一个关于求解一组正整数的最小公倍数（LCM）的问题，并提供了一段C语言实现代码示例。该问题要求输入一组正整数并返回这些数的最小公倍数。

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

2

    3 5 7 15 
   

    6 4 10296 936 1287 792 1
   

Sample Output

105

10296

Source

East Central North America 2003, Practice

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AC code:

#include " stdio.h "
__int64 s[ 1000 ];
__int64 hcf(__int64 a,__int64 b)
{
    __int64 r = 0 ;

     while (b != 0 )
    {
       r = a % b;
       a = b;
       b = r;
    }
     return (a);
}
__int64 lcd(__int64 u,__int64 v,__int64 h)
{
     return (u * v / h);
}

int main( int argc, char * argv[])
{
int n,m,i;
while (scanf( " %d " , & n) == 1 )
{
   while (n -- )
  {
   scanf( " %d " , & m);
    for (i = 0 ;i <= m - 1 ;i ++ )
    scanf( " %I64d " , & s[i]);
    for (i = 0 ;i <= m - 2 ;i ++ )
   {
    s[i + 1 ] = lcd(s[i],s[i + 1 ],hcf(s[i],s[i + 1 ]));
   }
   printf( " %I64d\n " ,s[m - 1 ]);

  }
}
return 0 ;

}

转载于:https://www.cnblogs.com/cpoint/archive/2011/04/13/2015298.html