我最初以为这是大树乘除法问题，后来发现只是普通的求最小公倍数方法。
原题：
Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38411 Accepted Submission(s): 14484
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
wa码
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#define max 100
using namespace std;
int cmp(int a,int b)
{
return b<a;
}
long long a[max],s;
int m,n,i,j;
int main()
{
scanf("%d",&n);
while(n--)
{ s=1;
<span style="white-space:pre"> </span>scanf("%d",&m);
for(i=0;i<m;i++)
scanf("%d",&a[i]);
sort(a,a+m,cmp);
for(i=0;i<m;i++)
{
for(j=i+1;j<m;j++)
{
if((a[i]%a[j])==0)
a[j]=1;
}
}
for(i=0;i<m;i++)
s=s*a[i];
printf("%d\n",s);
}
return 0;
}
总是出错，于是改啊改，发现算法就是错的，我的只能算15 5 7，正好谁是谁的公倍数的，却算不来哦36 8 14这种（此时虽然8 14不能被36 整除，但是他们存在最大公约数），要用我的算法，应该等于36*14*8，显然不对！
我的AC码
<span style="font-size:24px;">#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#define max 100
int main()
{
int a,n,m,i,s;
scanf("%d",&n);
while(n--)
{ s=0;
scanf("%d",&m);
while(m--)
{
scanf("%d",&a);
if(s==0)//</span><span style="font-size:14px;">先初始化让s西安等于第一个数的值</span><span style="font-size:24px;">
s=a;
for(i=s; ; i=i+s)//</span><span style="font-size:14px;">特别巧妙的算法，用i每次都乘以整数倍的本身，当i被a整除时恰好i是两者的最大公倍数
{</span><span style="font-size:24px;">
if(i%a==0)
break;
}
}
printf("%d",s);
}
return 0;
}
</span>
意外收获：
误以为qort（a，n，sizeof（a0），cmp）
int cmp (const void *a,const void *b)
可以用在c和c++都行！
原来这个是c语言的
c++应该是
qsort（a，a+n，cmp）
int cmp (int a,int b)
return a<b 或者a>b
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