题目链接

BZOJ2257

题解

由裴蜀定理我们知道，若干的瓶子如此倾倒最小能凑出的是其\(gcd\)

现在我们需要求出\(n\)个瓶子中选出\(K\)个使\(gcd\)最大

每个数求出因数排序即可

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<map>

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define mp(a,b) make_pair<int,int>(a,b)

#define cls(s) memset(s,0,sizeof(s))

#define cp pair<int,int>

#define LL long long int

using namespace std;

const int maxn = 1005,maxm = 1000005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int n,K;

int p[maxm],pi;

void sp(int x){

	p[++pi] = x;

	for (int i = 2; i * i <= x; i++)

		if (x % i == 0){

			p[++pi] = i;

			if (i * i != x) p[++pi] = x /  i;

		}

}

int main(){

	n = read(); K = read();

	REP(i,n) sp(read());

	sort(p + 1,p + 1 + pi);

	int ans = 1,cnt = 0;

	for (int i = 1; i <= pi; i++){

		if (p[i] != p[i - 1]){

			if (cnt >= K) ans = max(ans,p[i - 1]);

			cnt = 1;

		}

		else cnt++;

	}

	if (cnt >= K) ans = max(ans,p[pi]);

	printf("%d\n",ans);

	return 0;

}