【杭电oj2028】Lowest Common Multiple Plus

我不是女神ヾ 2022-09-25 06:16 97阅读 0赞

# Lowest Common Multiple Plus #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 52009 Accepted Submission(s): 21589**

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

Source

[C语言程序设计练习（五）][C]

Recommend

lcy | We have carefully selected several similar problems for you: [2046][] [1108][] [2044][] [2899][] [1215][]

[Statistic][] | [Submit][] | [Discuss][] | [Note][] 算是GCD

#include<stdio.h>
    #include<algorithm>
    using namespace std;
    int min(int a,int b);
    int main() {
        int n;
        while(scanf("%d",&n)==1) {
            int a[10000],k,ret;
            for(int l=0; l<n; l++)
                scanf("%d",&a[l]);
            sort(a,a+n);
            k=min(a[n-1],a[n-2]);
            for(int l=n-3; l>=0; l--) {
                ret=min(a[l],k);
                k=ret;
            }
            printf("%d\n",k);
        }
        return 0;
    }
    int min(int a,int b) {
        int i;
        for(int i=a; i<=a*b; i++)
            if(i%a==0&&i%b==0)
        return i;
    }

[C]: http://acm.hdu.edu.cn/search.php?field=problem&key=+C%D3%EF%D1%D4%B3%CC%D0%F2%C9%E8%BC%C6%C1%B7%CF%B0%A3%A8%CE%E5%A3%A9&source=1&searchmode=source
[2046]: http://acm.hdu.edu.cn/showproblem.php?pid=2046
[1108]: http://acm.hdu.edu.cn/showproblem.php?pid=1108
[2044]: http://acm.hdu.edu.cn/showproblem.php?pid=2044
[2899]: http://acm.hdu.edu.cn/showproblem.php?pid=2899
[1215]: http://acm.hdu.edu.cn/showproblem.php?pid=1215
[Statistic]: http://acm.hdu.edu.cn/statistic.php?pid=2028
[Submit]: http://acm.hdu.edu.cn/submit.php?pid=2028
[Discuss]: http://acm.hdu.edu.cn/discuss/problem/list.php?problemid=2028
[Note]: http://acm.hdu.edu.cn/note/note.php?pid=2028