POJ 3970（最小公倍數LCM）
阿新 • • 發佈：2017-06-03
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The CEO knows the number of employees in each of his teams and wants to determine X the least amount of money he should bring so that he awards the first team to show up such that all team members receive the same amount of money. You must write a program to help the CEO achieve this task.
知識點：
最小公倍數（a,b）=a*b/最大公約數（a。b）
Party
Description
The CEO of ACM (Association of Cryptographic Mavericks) organization has invited all of his teams to the annual all-hands meeting, being a very disciplined person, the CEO decided to give a money award to the first team that shows up to the meeting.The CEO knows the number of employees in each of his teams and wants to determine X the least amount of money he should bring so that he awards the first team to show up such that all team members receive the same amount of money. You must write a program to help the CEO achieve this task.
Input
The input consists of multiple test cases, each test case is described on a line by itself, Each line starts with an integer N (1 <= N <= 20) the number of teams in the organization followed by N space separated positive integers representing the number of employees in each of the N teams. You may assume that X will always fit in a 32 bit signed integer. The last line of input starts with 0 and shouldn‘t be processed.Output
For each test case in the input print "The CEO must bring X pounds.", where X is as described above or "Too much money to pay!" if X is 1000000 or more.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); //假設是int ，a*b將會溢出，造成錯誤
}
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;
}
POJ 3970（最小公倍數LCM）