首页 > 代码库 > ACM-简单题之Least Common Multiple——hdu1019

ACM-简单题之Least Common Multiple——hdu1019

2024-07-06 14:14:10   224人阅读

***************************************转载请注明出处：http://blog.csdn.net/lttree***************************************


Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28975    Accepted Submission(s): 10905


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


这道题就是求一堆数的最小公倍数。

要注意，每一行的第一个数表示后面跟着有几个数。

然后，我的方法是将输入的数存到数组，然后进行从大到小排序。

从头开始判断，如果当前的mul（之前所有数的最小公倍数）是当前数的倍数，就不需要计算这个数了，直接跳过。

最后再注意一点，用__int64来解决，scanf是用：%I64d


/****************************************
*****************************************
*        Author:Tree                    *
*From :  blog.csdn.net/lttree           *
* Title : Least Commn Multiple          *
*Source: hdu 1019                       *
* Hint :                                *
*****************************************
****************************************/

#include <stdio.h>
#include <algorithm>
using namespace std;

__int64 num[10001];

// 求a和b的最小公倍数
__int64 lcm( __int64 a,__int64 b)
{
    __int64 x=a,y=b,k;
    while( x%y!=0 )
    {
        k=x%y;
        x=y;
        y=k;
    }
    return a*b/k;
}
// 从大到小排序 sort用
bool cmp(__int64 a,__int64 b)
{
    return a>b;
}
int main()
{
    int t,n,i;
    __int64 mul;

    scanf("%d",&t);
    while( t-- )
    {
        scanf("%d",&n);
        for( i=0;i<n;++i )
            scanf("%I64d",&num[i]);
        sort(num,num+n,cmp);
        mul=num[0];
        // 如果当前的最小公倍数是该数的倍数就不需要计算了。
        for( i=1;i<n;++i )
            if( mul%num[i]!=0 )
                mul=lcm(mul,num[i]);
        printf("%I64d\n",mul);
    }
    return 0;
}


联系
我们