HDU 1019 数论 GCD和MCM 水题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37595    Accepted Submission(s): 14143

Problem Description

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.

 

 

Input

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.

 

 

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

 

 

Sample Input

2 3 5 7 15 6 4 10296 936 1287 792 1

 

 

Sample Output

105 10296

 

 

Source

East Central North America 2003, Practice

 

 

 

给出n个数，求出这n个数的LCM

 

刚开始看到这道题，因为没有给n的范围，觉得应该很大，估计是要分解各个数的。

 

 

后来暴力来一发，结果过了。

 

 

注意：就算运算结果在int内，运算过程也可能溢出。

 

 

 

 

 1 #include<cstdio>

 2 

 3 int gcd(int a,int b)

 4 {

 5     if(b==0)

 6         return a;

 7     return gcd(b,a%b);

 8 }

 9 

10 int main()

11 {

12     int test;

13     scanf("%d",&test);

14     while(test--)

15     {

16         int n;

17         scanf("%d",&n);

18         int x;

19         scanf("%d",&x);

20         for(int i=1;i<n;i++)

21         {

22             int tmp;

23             scanf("%d",&tmp);

24             x=(long long)x*tmp/gcd(x,tmp);

25         }

26         printf("%d\n",x);

27     }

28     return 0;

29 }

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