題意翻譯

給你n個數，去掉儘量少的數使得剩下數的gcd比原來的大。無解輸出-1

Translated by @xzz_233

題目描述

Mr. F has n n n positive integers, a1,a2,…,an a_1, a_2, \ldots, a_n a1​,a2​,…,an​ .

He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.

But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.

Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.

輸入輸出格式

輸入格式：

The first line contains an integer n n n ( 2≤n≤3⋅105 2 \leq n \leq 3 \cdot 10^5 2≤n≤3⋅105 ) — the number of integers Mr. F has.

The second line contains n n n integers, a1,a2,…,an a_1, a_2, \ldots, a_n a1​,a2​,…,an​ ( 1≤ai≤1.5⋅107 1 \leq a_i \leq 1.5 \cdot 10^7 1≤ai​≤1.5⋅107 ).

輸出格式：

Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.

You should not remove all of the integers.

If there is no solution, print «-1» (without quotes).

輸入輸出樣例

輸入樣例#1： 複製

3
1 2 4

輸出樣例#1： 複製

1

輸入樣例#2： 複製

4
6 9 15 30

輸出樣例#2： 複製

2

輸入樣例#3： 複製

3
1 1 1

輸出樣例#3： 複製

-1

說明

In the first example, the greatest common divisor is 1 1 1 in the beginning. You can remove 1 1 1 so that the greatest common divisor is enlarged to 2 2 2 . The answer is 1 1 1 .

In the second example, the greatest common divisor is 3 3 3 in the beginning. You can remove 6 6 6 and 9 9 9 so that the greatest common divisor is enlarged to 15 15 15 . There is no solution which removes only one integer. So the answer is 2 2 2 .

In the third example, there is no solution to enlarge the greatest common divisor. So the answer is −1 -1 −1 .

首先求出所有數的gcd，然後每個數都/gcd進行預處理，並統計每個數的個數；

此時我們要做的就是將此時  的序列move 若干個數，使得其 ；

考慮埃篩，我們篩質因數的時候將其倍數的個數也算上，最後 就是需要去掉的數的個數，那麼取 即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 500005
#define inf 0x3f3f3f3f
#define INF 999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

int n;
int a[maxn];
int Map[15000005];
int vis[15000005];

int main()
{
	//ios::sync_with_stdio(0);
	rdint(n);
	for (int i = 1; i <= n; i++)rdint(a[i]);
	int GCD = a[1];
	int Max = -inf;
	for (int i = 2; i <= n; i++)GCD = gcd(a[i], GCD);
	for (int i = 1; i <= n; i++) {
		Map[a[i] / GCD]++; Max = max(Max, a[i] / GCD);
	}
	int ans = 0;
	int tmp = inf;
	for (int i = 2; i <= Max; i++) {
		int res = 0;
		if (!vis[i]) {
			for (int j = i; j <= Max; j += i) {
				vis[j] = 1; res += Map[j];
			}
		}
		tmp = min(tmp, n - res);
	}
	if (tmp == inf)cout << -1 << endl;
	else cout << tmp << endl;
    return 0;
}

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