hdu 1019 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

这题就是一个简单求最小公约数的问题，使用欧几里得算法（辗转相除）求最小公约数；
然后利用公式：两数的乘积=最小公倍数*最大公约数；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
//辗转相除
int gcd(long long a,long long b)
{
if(a<b)
swap(a,b);
return b==0?a:gcd(b,a%b);
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int t;
scanf("%d",&t);
long long b=1,c;
int a;
while(t--)
{
scanf("%d",&a);
c=a*b;
b=c/gcd(a,b);
}
cout<<b<<endl;
}
return 0;
}

最后

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