ACM-简单题之Least Common Multiple——hdu1019 转载请注明出处：http://blog.csdn.net/lttree Least Common Multiple

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我是靠谱客的博主现代店员，最近开发中收集的这篇文章主要介绍ACM-简单题之Least Common Multiple——hdu1019 ***************************************转载请注明出处：http://blog.csdn.net/lttree*************************************** Least Common Multiple，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

转载请注明出处：http://blog.csdn.net/lttree

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28975 Accepted Submission(s): 10905

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

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1019

这道题就是求一堆数的最小公倍数。

要注意，每一行的第一个数表示后面跟着有几个数。

然后，我的方法是将输入的数存到数组，然后进行从大到小排序。

从头开始判断，如果当前的mul（之前所有数的最小公倍数）是当前数的倍数，就不需要计算这个数了，直接跳过。

最后再注意一点，用__int64来解决，scanf是用：%I64d

/****************************************
*****************************************
*
Author:Tree
*
*From :
blog.csdn.net/lttree
*
* Title : Least Commn Multiple
*
*Source: hdu 1019
*
* Hint :
*
*****************************************
****************************************/
#include <stdio.h>
#include <algorithm>
using namespace std;
__int64 num[10001];
// 求a和b的最小公倍数
__int64 lcm( __int64 a,__int64 b)
{
__int64 x=a,y=b,k;
while( x%y!=0 )
{
k=x%y;
x=y;
y=k;
}
return a*b/k;
}
// 从大到小排序 sort用
bool cmp(__int64 a,__int64 b)
{
return a>b;
}
int main()
{
int t,n,i;
__int64 mul;
scanf("%d",&t);
while( t-- )
{
scanf("%d",&n);
for( i=0;i<n;++i )
scanf("%I64d",&num[i]);
sort(num,num+n,cmp);
mul=num[0];
// 如果当前的最小公倍数是该数的倍数就不需要计算了。
for( i=1;i<n;++i )
if( mul%num[i]!=0 )
mul=lcm(mul,num[i]);
printf("%I64dn",mul);
}
return 0;
}