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=;i<n;i++)
         scanf("%d",&shuzu[i]);
         for(i=;i<n-;i++)
         {
             t=gcd(shuzu[i],shuzu[i+]);
             shuzu[i+]=shuzu[i]/t*shuzu[i+];
             //shuzu[i+1]=shuzu[i]*shuzu[i+1]/t;这样容易造成数据溢出
         }
         printf("%d\n",shuzu[n-]);
         free(shuzu);
     }
     //printf("%d\n",gcd(5,15));
     //while(1);
     return ;
 }
 //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=;i<n;i++)
         scanf("%d",&shuzu[i]);
         for(i=;i<n-;i++)
         for(j=shuzu[i];j<=shuzu[i]*shuzu[i+];j++)
         if(j%shuzu[i]==&&j%shuzu[i+]==)
         {
             shuzu[i+]=j;
             break;
         }
         printf("%d\n",shuzu[n-]);
         free(shuzu);
     }
     return ;
 }
 //TML