HDU1019

1019

Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53096 Accepted Submission(s): 20204

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

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**题目分析：
基本的求最小公倍数的方法是求出最大公约数，然后边输入边处理，注意数据长度，用的scanf和printf不知道怎么处理好unsigned long int型的数就用的更长的long long型**

#include<cstdio>int gcd(long long int a, long long b){    if (a%b == 0)return b;    else return gcd(b, a%b);}int lcm(long long a, long long b){    return a*b / gcd(a, b);}int main(){    int n,N;    scanf("%d", &N);    for (; N > 0; N--)    {        while (scanf("%d", &n) != EOF)        {            long long int temp1, temp2;            scanf("%lld", &temp1);            while (--n)            {                scanf("%lld", &temp2);                temp1 = lcm(temp1, temp2) ;            }            printf("%lld\n", temp1);        }    }}