HDU-ACM课堂作业 Least Common Multiple

Least Common Multiple

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 178   Accepted Submission(s) : 38

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.<br><br>

 

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.<br>

 

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.<br>

 

 

Sample Input

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

 

Sample Output

105 10296

 

Source

East Central North America 2003, Practice

 

这里INT-32类型的变量直接用int就可以啦

 

#include<iostream>
#include<stdio.h>
#include<algorithm> 
#include<string.h>
#include <iomanip>
#include<math.h>
using namespace std;

int GCD(int num1, int num2)
{
    if (num1%num2 == 0)
        return num2;
    else return  GCD(num2, num1%num2);
}
int LCM(int a, int b)
{
    int temp_lcm;
    temp_lcm = a / GCD(a, b)*b; //先除后乘，否则数据溢出导致WA
    return temp_lcm;
}

int main() {
    //数据组数
    int n = 0;
    while (cin >> n) {
        while (n--) {
            //变量数
            int k = 0;
            cin >> k;
            //将变量初始化
            int b = 0;
            int lcm, a = 1;
            for (int i = 0; i < k; i++) {
                //将每次输入的变量设置为b，计算a与b的LCM，随后将LCM赋值给a
                cin >> b;
                lcm = LCM(a, b);
                a = lcm;
            }
            cout << lcm << endl;
        }
    }
}