求大神指导,我到底错在哪!Problem DescriptionThe 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.InputInput

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求大神指导,我到底错在哪!Problem DescriptionThe 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.InputInput

求大神指导,我到底错在哪!Problem DescriptionThe 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.InputInput
求大神指导,我到底错在哪!
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
10510296
#include"stdio.h"
int main()
{
int n,i,j,k,m,a[1000],min;
scanf("%d",&n);
for(i=1;i=1;k--)
if(a[0]%k==0&&a[1]%k==0)
break;
min=(a[0]*a[1])/k;
for(j=2;ja[j])
{
k=min;
min=a[j];
a[j]=k;
}
for(k=min;k>=1;k--)
if(min%k==0&&a[j]%k==0)
break;
min=(min*a[j])/k;
}
printf("%d\n",min);
}
return 0;
}

求大神指导,我到底错在哪!Problem DescriptionThe 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.InputInput
问题太多 完全达不到题目要求的意思
题目看懂了没?就是求最小公倍数 所有输入数字的
不给你改了 你看下这个
#include
#include
long calc_lcm(long a, long b)
{
long i, min;
min = a 0; i --)
if((a % i == 0) && (b % i == 0)) break;
return a*b/i;
}
int main()
{
int loop;
long *array;
int size, i;
long lcm;

printf("input loop:");
scanf("%d", &loop);
while(loop --)
{
printf("please input number count and numbers\n");
scanf("%d", &size);

if(size