hdu 1019 Least Common Multiple

最新推荐文章于 2025-09-12 18:08:49 发布

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原文链接：http://www.cnblogs.com/superxuezhazha/p/5539794.html

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#java

本文介绍了一种求解多个正整数最小公倍数（LCM）的有效算法，并提供了完整的C++代码实现。该算法通过计算每对数的最大公约数（GCD），进而求得最小公倍数。

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44922 Accepted Submission(s): 16877

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

直接算就可以了，没输入一个数求当前数和前两个数的最小公倍数的最小公倍数。

#include<iostream>
#include<stdio.h>
using namespace std;
long long gcd(long long x,long long y)
{
    long long a;
    if(x<y) a=x,x=y,y=a;
    while(y!=0)
    {
        a=x%y;
        x=y;
        y=a;
    }
    return x;
}
int main()
{
    int t,n;
    long long tmp,end;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        end=1;
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&tmp);
            end=(tmp*end)/(gcd(tmp,end));
        }
        printf("%I64d\n",end);
    }
    return 0;
}