hdoj 1019 最小公倍数问题

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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

23 5 7 156 4 10296 936 1287 792 1

Sample Output

10510296

wa码

#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<string.h>#define max 100using 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

#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<string.h>#define max 100int 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)//先初始化让s先等于第一个数的值 s=a;for(i=s; ; i=i+s)//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 if(i%a==0)break;}}printf("%d",s);}return 0;}

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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