计算n个数的最小公倍数,可用欧几里得算法计算两个数字的最大公约数,再计算两个数最小公倍数
有了2个数最小公倍数算法就简单了,即为:计算第一和第二个数得到最小公倍数lc,再计算lc和第三个数最小公倍数......
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
inline int lcm(int a,int b);
int main()
{
int N, n,a, i;
cin >> N;
while (N--)
{
cin >> n;
vector<int>v;
for (i = 0; i < n&&cin >> a; i++)
v.push_back(a);
if (n == 1)
{
cout << v[0] << endl;
continue;
}
int lc = lcm(v[0], v[1]);
for (i=2; i < n; i++)
lc = lcm(lc, v[i]);
cout << lc << endl;
}
return 0;
}
inline int lcm(int a, int b)
{
int r,ra=a,rb=b; //ra,rb记录初始传入是的a,b的值
if (a<b)
swap(a, b);
while (a%b!= 0)
{
r = a%b;
a = b;
b = r;
}
int lc = ra*(rb / b);
return lc;
}
时间: 2024-10-12 22:15:21