HDU

Least Common Multiple

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51937 Accepted Submission(s): 19700

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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.

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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.

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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.

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Sample Input

2
3 5 7 15
6 4 10296 936 1287 792 1

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Sample Output

105
10296

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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 longll 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