Least Common Multiple

问题 I: Least Common Multiple

时间限制: 1 Sec  内存限制: 32 MB
提交: 40  解决: 16
[提交][状态][讨论版]

题目描述

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.

输入

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.

输出

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

样例输入

22 3 53 4 6 12

样例输出

1512

提示

解题思路：两个数的最小公倍数等于两个数的最大公约其中一个数除以最大公约数然后乘以另一个数。为了防止溢出，先除后乘。

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std; int a[10010];int comm(int a, int b){if(b==0)return a;return comm(b,a%b);}int main(){int sum, t, n,i,flag;scanf("%d", &t);while(t--){scanf("%d", &n);for(i=0; i<n; i++)scanf("%d", a+i);sum=1;for(i=0,flag=0; i<n-1; i++){flag=comm(a[i],a[i+1]);a[i+1]=((a[i]/flag)*(a[i+1]/flag)*flag);}printf("%d\n", a[n-1]);}}