hdu1019 Least Common MultipleLeast Common Multiple

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概述

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 59825 Accepted Submission(s): 22822

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

求多个数的最小公倍数

#include<stdio.h>
#include<algorithm>
using namespace std;
int LCM(int m,int n)
{
int a,b,c;
a=m,b=n;
while(b!=0)
{
c=a%b;
a=b;
b=c;
}
return m/a*n;//防止溢出
}
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int num[n];
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
if(n==1) //注意n为1的特殊情况
{
printf("%dn",num[0]);
}
else
{
sort(num,num+n);
for(int i=0;i<n-1;i++)
{
num[i+1]=LCM(num[i],num[i+1]);
}
printf("%dn",num[n-1]);
}
}
return 0;
}