Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33702    Accepted Submission(s): 13766

Problem Description

求n个数的最小公倍数。

 

Input

输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。

 

Output

为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。

 

Sample Input

2 4 6
3 2 5 7

 

Sample Output

12
70

 

Author

lcy

 

Source

C语言程序设计练习（五）

 

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Statistic | Submit | Discuss | Note

正常的水题吧。

先用简单的辗转相除法求出最大公约数，再求出最小公倍数。

注意需要使用unsigned才可AC。

#include <iostream>
using namespace std;
unsigned getcount(unsigned a, unsigned b){
	unsigned c;
	unsigned m = a;
	unsigned n = b;
	if (a < b){
		unsigned temp = a;
		a = b;
		b = temp;
	}
	while (b){
		c = a%b;
		a = b;
		b = c;
	}
	return (m*n) / a;
}
int main(){
	unsigned n, res, temp;
	while (cin >> n){
		if (n > 0){
			cin >> temp;
			res = temp;
			n--;
		}
		while (n--){
			cin >> temp;
			res = getcount(res, temp);
		}
		cout << res << endl;
	}
	return 0;
}

HDOJ 2028 Lowest Common Multiple Plus