Lowest Common Multiple Plus

Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42356    Accepted Submission(s): 17505

Problem Description

求n个数的最小公倍数。

 

Input

输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。

 

Output

为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。

 

Sample Input

2 4 6 3 2 5 7

 

Sample Output

12 70

 

Author

lcy

#include<stdio.h>
int main (void)
{
    int a[500];
    int i, n, temp;
    while(scanf("%d", &n) == 1){
        temp = 0;
        for(i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            if(temp < a[i])
                temp = a[i]; 
        }
        for(i = 1; i <= n; i++)
            if(temp % a[i] != 0){
                temp++;
                i = 0; 
            }
        printf("%d\n", temp);
    }
    return 0;
}