hdu 1019 Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52181    Accepted Submission(s): 19805

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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296
题目解析：
这个题就是让求几个数的最小公倍数，首先我们要对数列进行排序，找到最大值，开始向上增大，每次增大最大的数，用循环就可以找到最小公倍数
代码：
#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;int main(){    int t,s,n,i,j,m,a[5000];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)            scanf("%d",&a[i]);        sort(a,a+n);        for(i=a[n-1];;i+=a[n-1])        {            for(j=0;j<n-1;j++)                if(i%a[j]!=0)                    break;            if(j==n-1)                break;        }        printf("%d\n",i);    }    return 0;}