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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.
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
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。
从头開始推断，假设当前的mul（之前全部数的最小公倍数）是当前数的倍数，就不须要计算这个数了，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];
// 假设当前的最小公倍数是该数的倍数就不须要计算了。
for( i=1;i<n;++i )
if( mul%num[i]!=0 )
mul=lcm(mul,num[i]);
printf("%I64d\n",mul);
}
return 0;
}