CodeForces 389A Fox and Number Game

链接：http://codeforces.com/problemset/problem/389/A

Fox and Number Game

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Fox Ciel is playing a game with numbers now.

Ciel has n positive integers: x₁, x₂, ..., x_n. She can do the following operation as many times as needed: select two different indexes i and j such that x_i > x_j hold, and then apply assignment x_i = x_i - x_j. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.

Input

The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x₁, x₂, ..., x_n (1 ≤ x_i ≤ 100).

Output

Output a single integer — the required minimal sum.

Sample test(s)

Input

2
1 2

Output

2

Input

3
2 4 6

Output

6

Input

2
12 18

Output

12

Input

5
45 12 27 30 18

Output

15

Note

In the first example the optimal way is to do the assignment: x₂ = x₂ - x₁.

In the second example the optimal sequence of operations is: x₃ = x₃ - x₂, x₂ = x₂ - x₁.

大意——有一个游戏规则是这样的：给你n个正整数，从中找到两个不同的数，将较大的数变成大数减去小数的差。你可以进行多次操作，目标是使得最后得到的n个数的和尽量小。

思路——一道含数学问题的贪心题。由数学中的更相减损术可以知道，两个数一直作差运算，最后一定是相等的，并且最后这个数为原来的两个数的最大公约数。那么问题中的n个数要使其和最小，就必须作很多次差运算，直至最后所有数全部相等。那么就可以知道最后这个数为这n个数的最大公约数，从而问题转化为求这n个数的最大公约数，最后将它们的最大公约数乘以n即为答案。

复杂度分析——时间复杂度:O(n),空间复杂度:O(1)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
int n, x;

int gcd(int a, int b);

int main()
{
	ios::sync_with_stdio(false);
	while (~scanf("%d", &n))
	{
		int ans=0, cnt=n;
		while (n--)
		{
			scanf("%d", &x);
			ans = gcd(ans, x);
		}
		printf("%d\n", ans*cnt);
	}
	return 0;
}

int gcd(int a, int b)
{
	return b ? gcd(b, a%b) : a;
}