HDOJ 2028 Lowest Common Multiple Plus

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個數的最小公倍數。
Input 輸入包含多個測試實例，每個測試實例的開始是一個正整數n，然後是n個正整數。
Output 為每組測試數據輸出它們的最小公倍數，每個測試實例的輸出占一行。你可以假設最後的輸出是一個32位的整數。
Sample Input

2 4 6
3 2 5 7

Sample Output

12
70

Author lcy
Source C語言程序設計練習（五）
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Statistic | Submit | Discuss | Note

正常的水題吧。

先用簡單的輾轉相除法求出最大公約數，再求出最小公倍數。

注意需要使用unsigned才可AC。

#include 
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;
}