HDU 1019 数论 GCD和MCM 水题

LCM求解算法实现

最新推荐文章于 2016-08-17 22:50:38 发布

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原文链接：http://www.cnblogs.com/-maybe/p/4523540.html

本文介绍了一种求解多个正整数最小公倍数（LCM）的算法实现，包括输入格式解析、LCM计算过程及输出结果展示。

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37595 Accepted Submission(s): 14143

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

给出n个数，求出这n个数的LCM

刚开始看到这道题，因为没有给n的范围，觉得应该很大，估计是要分解各个数的。

后来暴力来一发，结果过了。

注意：就算运算结果在int内，运算过程也可能溢出。

 1 #include<cstdio>
 2 
 3 int gcd(int a,int b)
 4 {
 5     if(b==0)
 6         return a;
 7     return gcd(b,a%b);
 8 }
 9 
10 int main()
11 {
12     int test;
13     scanf("%d",&test);
14     while(test--)
15     {
16         int n;
17         scanf("%d",&n);
18         int x;
19         scanf("%d",&x);
20         for(int i=1;i<n;i++)
21         {
22             int tmp;
23             scanf("%d",&tmp);
24             x=(long long)x*tmp/gcd(x,tmp);
25         }
26         printf("%d\n",x);
27     }
28     return 0;
29 }