HDOJ 1019 Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23664    Accepted Submission(s): 8847

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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296

 

这是  WA代码  

#include <stdio.h>#include <iostream>using namespace std;int gcd(int a,int b){    return b==0? a:gcd(b,a%b);}int main(){    int n,a,sum,i,a1[100005];    scanf("%d",&n);    while (n--)    {        scanf("%d",&a);        for (i=0;i<a;i++)        {            scanf("%d",&a1[i]);        }    sum=a1[0];    for (i=1;i<a;i++)    {        sum=(sum*a1[i])/(gcd(a1[i],sum));    }    printf("%d\n",sum);    }    return 0;}

这个是AC代码  真心觉得就是求最小公倍数的算法不同 但是最小公倍数不是拿两数之积除以最大公约数么 可是WA了  换了一直算法就对了

#include <stdio.h>
#include <iostream>
using namespace std;
int gcd(int a,int b)
{
    return b==0? a:gcd(b,a%b);
}
int main()
{
    int n,a,sum,i,a1[100005];
    scanf("%d",&n);
    while (n--)
    {
        scanf("%d",&a);
        for (i=0;i<a;i++)
        {
            scanf("%d",&a1[i]);
        }
    sum=a1[0];
    for (i=1;i<a;i++)
    {
        sum=sum*(a1[i]/gcd(a1[i],sum));
    }
    printf("%d\n",sum);
    }
    return 0;
}