hdu_1019_Least Common Multiple_201310290920

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24649    Accepted Submission(s): 9281

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

 

题意：求几个数的最小公倍数

思路:前两个数求一次，求得的结果再与下一个数求

 

//hdu_1019_Least Common Multiple_201310290920-2
#include <stdio.h>
#include <malloc.h>

void swap(int* a,int* b)
{
    int t;
    t=*a;
    *a=*b;
    *b=t;   
}
int gcd(int m,int n)
{
    int i;
    if(m>n)
    swap(&m,&n);
    i=m;
    while(i)
    {
        i=n%m;
        n=m;
        m=i;
    }
    return n;
}
int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        int i,j,n,t;
        int *shuzu;
        scanf("%d",&n);
        shuzu = (int*)malloc(sizeof(int)*n);
        for(i=0;i<n;i++)
        scanf("%d",&shuzu[i]);
        for(i=0;i<n-1;i++)
        {
            t=gcd(shuzu[i],shuzu[i+1]);
            shuzu[i+1]=shuzu[i]/t*shuzu[i+1];
            //shuzu[i+1]=shuzu[i]*shuzu[i+1]/t;这样容易造成数据溢出 
        }
        printf("%d
",shuzu[n-1]);
        free(shuzu);
    }
    //printf("%d
",gcd(5,15));
    //while(1);
    return 0;
}
//ac

#include <stdio.h>
#include <malloc.h>

int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        int i,j,n;
        int *shuzu;
        scanf("%d",&n);
        shuzu = (int*)malloc(sizeof(int)*n);
        for(i=0;i<n;i++)
        scanf("%d",&shuzu[i]);
        for(i=0;i<n-1;i++)
        for(j=shuzu[i];j<=shuzu[i]*shuzu[i+1];j++)
        if(j%shuzu[i]==0&&j%shuzu[i+1]==0)
        {
            shuzu[i+1]=j;
            break;
        }
        printf("%d
",shuzu[n-1]);
        free(shuzu);
    }
    return 0;
}
//TML