2028

Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 59375    Accepted Submission(s): 24649

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 63 2 5 7

 

Sample Output

1270

 

Author

lcy

 

Source

C语言程序设计练习（五）

 

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View Code

Problem : 2028 ( Lowest Common Multiple Plus )     Judge Status : Accepted
RunId : 20106356    Language : C    Author : 837274600
Code Render Status : Rendered By HDOJ C Code Render Version 0.01 Beta

#include<stdio.h>#include<string.h>#include<math.h>int main(){    int n;    while (scanf("%d", &n) != EOF)    {        int a[100], i, max = 0;        for (i = 0; i < n; i++)        {            scanf("%d", &a[i]);            if (a[i] > max)                max = a[i];        }        int flag = 1, m = max;        while (flag)        {            for (i = -1; i++ < n - 1; )            {                if (m%a[i] != 0)                    break;            }            if (i == n)                flag = 0;            else                m += max;        }        printf("%d\n", m);    }    return 0;}