Least Common Multiple&&http://acm.hdu.edu.cn/showproblem.php?pid=1019

最新推荐文章于 2024-08-29 11:58:30 发布

原创最新推荐文章于 2024-08-29 11:58:30 发布 · 810 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#integer #output #numbers #input #each #n2

HDU 同时被 2 个专栏收录

127 篇文章

订阅专栏

数论

64 篇文章

订阅专栏

本文介绍了一种求解一组正整数最小公倍数的有效算法，通过质因数分解并利用快速幂运算提高效率，适用于编程竞赛及数学问题解决。

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
题意：给你一些数，求它们的最小公倍数，，，这是一道数论题，本以为过不了的，没想到数据那么水~~~
AC代码：
#include<iostream>
#include<string.h>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cmath>
#define N 1005
using namespace std;
int prim[N];
bool s[1000005]={1,1,0};
int num[N];
int res;
void init()
{
	
	memset(num,0,sizeof(num));
}
void _prim()
{
	res=0;
	memset(prim,0,sizeof(prim));
	int m=sqrt(1000005+0.5);
	for(int i=2;i<=m;++i)
		if(!s[i])
		{
			prim[res++]=i;
			for(int j=i*i;j<=1000005;j+=i)
				s[j]=1;
		}
}
int _pow(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1) ans*=a;
		 a=a*a;
		 b=b>>1;
	}
	return ans;
}
void in(int &a)
{
	char ch;
	while((ch=getchar())<'0'||ch>'9');
	for(a=0;ch>='0'&&ch<='9';ch=getchar()) a=a*10+ch-'0';
}
void out(int x)
{
	if(x>9) out(x/10);
	putchar(x%10+48);
}
int main()
{
	int T;
	in(T);
	_prim();
	while(T--)
	{
		init();
		int n;
		in(n);
		for(int i=0;i!=n;++i)
		{
			int a;
			in(a);
			int ans=a;
			for(int i=0;i<res;++i)
			{
				if(ans%prim[i]==0)
			   {
				   int cnt=0;
				   while(ans%prim[i]==0)
				  {
					 cnt++;
					 ans/=prim[i];
				   }
				   num[prim[i]]=max(num[prim[i]],cnt);
				   //cout<<prim[i]<<" "<<num[prim[i]]<<endl;
			    }
			 if(ans==1) break;
			}
			//if (ans>1) num[ans]=max(num[ans],1);
		}
		int sum=1;
		for(int i=0;i<res;++i)
			if(num[prim[i]])  sum=sum*_pow(prim[i],num[prim[i]]);
		out(sum);
		printf("\n");
	}return 0;
}