1019 

 1 
 2  /*Description: to 
 3  * 
 4  * Author: Vincent
 5  * 
 6  * Date:2010/04/
 7  * Contact:agilely@126.com 
 8  * Zju CS Lab
 9 
10 
11 
12 
13   Problem Description
14 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.
15 
16 
17  
18 
19 Input
20 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.
21 
22  
23 
24 Output
25 For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
26 
27  
28 
29 Sample Input
30 2
31 3 5 7 15
32 6 4 10296 936 1287 792 1
33  
34 
35 Sample Output
36 105
37 10296
38  
39 
40 Source
41 East Central North America 2003, Practice 
42 
43 
44 
45  */
46 //1019多个数的最小公倍数
47 //先把数按从大到小排序，再两个两个求
48 #include <iostream>
49 using namespace std;
50 long  LCM(long  num1,long  num2)
51 {
52     long int a,b,temp;
53     if(num1>0 && num2>0)
54     {
55         if(num1<num2)/*交换两个数，使大数放在num1上*/
56         {
57             temp=num1;
58             num1=num2;
59             num2=temp;
60         }
61         a=num1;b=num2;
62         while(b!=0)/*利用辗除法，直到b为0为止*/
63         {
64             temp=a%b;
65             a=b;
66             b=temp;
67         }
68         num1/=a;
69         return num1*num2;
70     }
71     else
72     {  return 0;}
73 }
74 int main()
75 {
76     int t,n,ca,cb;
77     long num1,num2;
78     if(cin>>t)//用例数
79     {
80         for(ca=0;ca<t;ca++)
81         {
82             cin>>n;//每个用例中数的个数
83             cin>>num1;//第一个数
84             for(cb=1;cb<n;cb++)
85             {
86                 cin>>num2;//从第二个数开始
87                 num1=LCM(num1,num2);
88             }//n
89             cout<<num1<<endl;
90         }//t
91     }
92     return 0;
93 }
94 
95

 

posted @ 2010-04-15 23:15  にんじゃ  阅读(136)  评论(0)    收藏  举报