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。

 

 

 

 

#include<cstdio>

int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}

int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        int n;
        scanf("%d",&n);
        int x;
        scanf("%d",&x);
        for(int i=1;i<n;i++)
        {
            int tmp;
            scanf("%d",&tmp);
            x=(long long)x*tmp/gcd(x,tmp);
        }
        printf("%d
",x);
    }
    return 0;
}