HDU - 1019 Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 51937 Accepted Submission(s): 19700

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

题目大意就是找出n个数的最小公倍数，然后那个32 - bit 应该是长整型（long long）

主要的做题思路就是不断的求最小公倍数

#include<iostream>
#include<stdio.h>
using namespace std;
#define ll long long
ll gcd (ll a ,ll b )//辗转相除（迭代法）
{
while( a>0 )
{
ll c = b % a;
b = a ;
a = c ;
}
return b ;
}
int main()
{
int t;
cin>>t;
while(t--)
{
ll n;
scanf("%lld",&n);
ll ans = 1;
while(n--)
{
ll m;
scanf("%lld",&m);
if(ans%m==0)
continue;
ll g=gcd(ans,m);
ans = ans*m/g; //这里用到了gcd (a,b)*lcm(a,b)=a*b
}
printf("%lld\n",ans);
}
return 0;
}

下面是整理出的一些求GCD（最大公因数）的方法：

1 . 递归法：欧几里得算法 -> 辗转相除法 gcd( a , b ) = gcd( b , a mod b )

int gcd ( int a , int b)
{
//if( a > b )  swap( a , b )
return (b==0) ? a : gcd ( b , a % b );
}

2 . 迭代法（递推法）：欧几里得算法 ///// 较快

int gcd ( int a , int b )
{
while( a>0 )
{
int c = b % a;
b = a ;
a = c ;
}
return b ;
}

3 . 连续整数试探法 // 就是最为普通的遍历

int gcd ( int a , int b )
{
if( a > b )    // 使 a < b
{
int temp = a ;
a = b ;
b = temp ;
}
int t = a ;
while( a % t || b % t)
{
t - - ;
}
return t ;
}

LCM（最小公倍数）

基于GCD 由 gcd ( a , b ) × lcm( a , b ) = a × b 得到 LCM