hdu 1019 Least Common Multiple

Least Common Multiple

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 4

Font: Times New Roman | Verdana |Georgia

Font Size: ← →

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

 

 

 

 

代码#include <stdio.h>
int max(int a,int b)
{
 int t;
 if (a<b)
 {
  t=a;a=b;b=t;
 }
 while (b!=0)
 {
  t=a%b;
  a=b;
  b=t;
 }
 return a;
}
int main()
{
 int M,N,i,j,a[100],min;
 scanf("%d",&M);    //注意M，N各自控制的是什么
 while (M--)
 {
  scanf("%d",&N);
  for (i=1;i<=N;i++)
  {
   scanf("%d",&a[i]);
  }
  for (i=1;i<N;i++)      
  {
   min=a[i]/max( a[i], a[i+1])]*a[i+1];   // //如果先乘法后除法，乘法会溢出，WA
   a[i+1]=min;
  }
  printf("%d\n",a[N]);

 }
 return 0;
}