POJ 3970（最小公倍数LCM）

知识点：

     最小公倍数（a,b）=a*b/最大公约数（a。b）

                                                                                                           Party

Description

Input

Output

Sample Input

1 3000000
2 12 4
0

Sample Output

Too much money to pay!
The CEO must bring 12 pounds.

#include<iostream>
#include<cstdio>

using namespace std;

__int64 num[30];

__int64 gcd(__int64 a,__int64 b)
{
    __int64 r;
    __int64 t;
    if(a<b)
    {
        t=a;
        a=b;
        b=t;
    }
    r=a%b;
    while(r)
    {
        a=b;
        b=r;
        r=a%b;
    }
    return b;
}

__int64 lcm(__int64 a,__int64 b)
{
    return a*b/gcd(a,b);        //假设是int ，a*b将会溢出，造成错误
}


int main()
{
   int t;
   __int64 res;
   while(scanf("%d",&t),t)
   {
       for(int i=0;i<t;i++)
        scanf("%I64d",num+i);
       res=num[0];
       //cout<<gcd(num[0],num[0]);

       for(int i=1;i<t;i++)
        res=lcm(num[i],res);

       if(res>=1000000)
           printf("Too much money to pay!\n");
        else
            printf("The CEO must bring %I64d pounds.\n",res);
   }
   return 0;
}