【51Nod 1616】【算法马拉松 19B】最小集合

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1616
这道题主要是查询一个数是不是原有集合的一个子集的所有数的gcd。
只要枚举一个数的倍数暴力判断就可以了，时间复杂度\(O(alna),a=10^6\)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

bool vis[1000003];

int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}

int main() {
	int n, x, num, ans = 0, top = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &x), vis[x] = true, top = max(top, x);
	
	for (int i = 1; i <= top; ++i)
		if (!vis[i]) {
			num = -1;
			for (int j = (i << 1); j <= top; j += i)
				if (vis[j]) {
					if (num == -1)
						num = j;
					else
						num = gcd(num, j);
					if (num == i)
						vis[i] = true;
				}
		}
	
	for (int i = 1; i <= top; ++i)
		if (vis[i])
			++ans;
	printf("%d\n", ans);
	return 0;
}