专栏首页小樱的经验随笔HDU 1019 Least Common Multiple【gcd+lcm+水+多个数的lcm】

HDU 1019 Least Common Multiple【gcd+lcm+水+多个数的lcm】

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 53016    Accepted Submission(s): 20171

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

2
3 5 7 15
6 4 10296 936 1287 792 1

Sample Output

105
10296

Source

East Central North America 2003, Practice

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1019

分析:多个数的最小公倍数,直接每次对两个进行lcm操作就好了,具体代码如下:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 inline ll gcd(ll a,ll b)
 5 {
 6     return b==0?a:gcd(b,a%b);
 7 }
 8 inline ll lcm(ll a,ll b)
 9 {
10     return a*b/gcd(a,b);
11 }
12 ll a[100010];
13 int main()
14 {
15     ll n;
16     while(scanf("%lld",&n)!=EOF)
17     {
18         while(n--)
19         {
20             ll m;
21             scanf("%lld",&m);
22             for(ll i=1;i<=m;i++)
23                 scanf("%lld",&a[i]);
24             for(ll i=1;i<=m-1;i++)
25                 a[i+1]=lcm(a[i],a[i+1]);
26             printf("%lld\n",a[m]);
27         }
28     }
29     return 0;
30 }

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

我来说两句

0 条评论
登录 后参与评论

相关文章

  • Codeforces 768B Code For 1

    B. Code For 1 time limit per test:2 seconds memory limit per test:256 megabytes ...

    Angel_Kitty
  • Codeforces 810C Do you want a date?(数学,前缀和)

    C. Do you want a date? time limit per test:2 seconds memory limit per test:256 m...

    Angel_Kitty
  • Codeforces Round #395 (Div. 2)(A.思维,B,水)

    A. Taymyr is calling you time limit per test:1 second memory limit per test:256 ...

    Angel_Kitty
  • Codeforces Round #619 (Div. 2)

    A 题目 You are given three strings a, b and c of the same length n. The strings ...

    杨鹏伟
  • PAT (Advanced Level) Practice 1146 Topological Order (25分)

    This is a problem given in the Graduate Entrance Exam in 2018: Which of the foll...

    glm233
  • 【Python机器学习】支持向量机(附源码)

    从本周开始,推送一个系列关于 Python 机器学习 。为了保证内容的原汁原味。我们采取全英的推送。希望大家有所收获。提高自己的英语阅读能力和研究水平。 In...

    量化投资与机器学习微信公众号
  • 代码转换工具 Code Converter 2013

    http://www.codeproject.com/Articles/260997/Code_Converter This is a Code Conver...

    张善友
  • 用树状结构解码复杂类别的长尾进行超标记(CS CL)

    虽然目前的CCG超级测试器在标准的WSJ测试集上达到了很高的准确率,但很少有系统利用类别的内部结构来驱动句法分析过程中的句法推导。标记集被截断,丢弃了在长尾中许...

    谭雪儿
  • ZOJ 3203 Light Bulb

    Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. H...

    ShenduCC
  • Codeforces Beta Round #14 (Div. 2)A. Letter

    A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sh...

    glm233

扫码关注云+社区

领取腾讯云代金券