HDU 1019

最新推荐文章于 2018-12-03 18:49:57 发布

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本文介绍了一种求多个正整数最小公倍数(LCM)的算法，并提供了一个具体的C++实现示例。该算法适用于东中北美洲2003年练习赛题目，通过逐对计算最大公因数来确定最小公倍数。

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

//就是求几个数的最小公倍数。输出格式要注意，第一个数是样例数，第二个数是单个样例输入个数。

//读了半天才弄懂。各种借鉴大神，最后修改类型才通过！

#include <iostream>

#include <cstdio>

using namespace std;

__int64 LCM(__int64 a,__int64 b){//求最大公倍数
    __int64 c;
    __int64 a1=a,b1=b;
    if(a<b){
        c=b;
        b=a;
        a=c;
    }
    while(b){
        c=a%b;
        a=b;
        b=c;
    }
    return (a1*b1)/a;
}

int main()
{
   int n,T;
   __int64 a[10000],result;
   scanf("%d",&T);
   while(T--){
        scanf("%d",&n);
        for(__int64 i=0;i<n;i++)
            scanf("%d",&a[i]);
        if(n==1){
            printf("%I64d\n",a[0]);
        }else{
        result=LCM(a[0],a[1]);
        for(__int64 i=2;i<n;i++)
            result=LCM(result,a[i]);
        printf("%I64d\n",result);
        }
       }
    return 0;
}