HDU 1019 Least Common Multiple

Least CommonMultiple

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44190    Accepted Submission(s): 16580

Problem Description

The least common multiple (LCM) of a set of positive integers is thesmallest positive integer which is divisible by all the numbers in the set. Forexample, the LCM of 5, 7 and 15 is 105.

 

 

Input

Input will consist of multipleproblem instances. The first line of the input will contain a single integer indicatingthe number of problem instances. Each instance will consist of a single line ofthe 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 rangeof a 32-bit integer.

 

 

Output

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

 

 

Sample Input

2

3 5 7 15

6 4 10296 936 1287 792 1

 

 

Sample Output

105

10296

 

题目就是求最小公倍数，每次输入一个数后都记录下当前最小公倍数，虽然一次AC，但是运行用时多，以后多做改进

#include<stdio.h>int main(){int n,m,a,k;while(scanf("%d",&n)!=EOF&&n){while(n--){scanf("%d",&m);int t=1;while(m--){int i=2;scanf("%d",&a);k=a;//记录下a的原值 while(a%t!=0)//求出最小公倍数 {a=k*i;i++;} t=a;//记录下当前最小公倍数a }printf("%d\n",a);}return 0;}}