n个数的最小公倍数

求n个数的最小公倍数 比如 1 2 3 12
先求1 2 的最小公倍数m
然后求 m 3的最小公倍数 m
再求m 12.。。。 循环

ll mid = lcm(arr[i], lcm(arr[j], lcm(arr[l],arr[k])));   求4个数的最小公倍数。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std ;
#define  ll long long
const int M = 51;
int n,t;
ll arr[M];
ll gcd(ll a, ll b)
{
    if(b == 0) return a;
    else return gcd(b, a%b);
}
ll lcm(ll a, ll b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    while(scanf("%d%d",&n,&t))
    {
        if(n < 4) break;
        if(!n || !t) break;
        for(int i = 0; i < n; ++i)
        scanf("%d",&arr[i]);

        ll h;
        while(t--)
        {
            scanf("%d",&h);
            int flag = 0;
            ll mmin = 1000000000;
            ll a , b = 1000000000;
            for(int i = 0; i < n-3; i++)
            {
                for(int j = i+1; j < n-2; j++)
                {
                    for(int l = j+1; l < n-1; l++)
                    {
                        for(int k = l+1; k < n; k++)
                        {
                            ll mid = lcm(arr[i], lcm(arr[j], lcm(arr[l],arr[k])));

                            ll temp = h % mid;
                            a = mid - temp;
                            if(temp == 0)
                            {
                                b = 0;
                                mmin = 0;
                                flag  =1 ;break;
                            }
                            else
                            {
                                if(temp < mmin) mmin = temp;
                                if(a < b) b = a;
                            }
                        }
                        if(flag) break;
                    }
                    if(flag) break;
                }
                if(flag) break;
            }
            printf("%d %d\n",h-mmin,h+b);
        }
    }