【Codeforces】1047 - C. Enlarge GCD

题目链接

题意：给出n个数字，求最少删除几个可以使剩下的数字的GCD大于n个数字的GCD。

思路：求出最多的公共因子，去掉没有这个因子的数，剩下的数的GCD大于原来的GCD。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define clr(s, x) memset(s, x, sizeof(s))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
inline int read(){int r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll readll(){ll r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll qpow(ll a,ll b,ll mod){ll res=1;while(b){if(b&1)res = (res*a)%mod;a=(a*a)%mod;b>>=1;}return res;}
inline ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const double eps = 1e-8;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 1.5e7+5;
const int MAXM = 1.5e7+5;
int Mark[MAXN];  
int prime[MAXN];  

int Prime(){  
    int tot = 0;  
    memset(Mark, 0, sizeof(Mark));  
    Mark[0] = 1; Mark[1] = 1;
    for(int i = 2; i < MAXN; i++)  
    {  
        if(Mark[i] == 0){  
            prime[tot++] = i;  
        }  
        for(int j = 0; j < tot && prime[j] * i < MAXN; j++)  
        {  
            Mark[i * prime[j]] = 1;  
            if(i % prime[j] == 0){  
                break;  
            }  
        }  
    }  
    return tot;  
}  

int a[300005], cnt[MAXN];
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int tot = Prime();
	int n;
	scanf("%d", &n);
	int g = 0;
	for(int i=0; i<n; ++i){
		scanf("%d", &a[i]);
		g = __gcd(g, a[i]);
	}
	for(int i=0; i<n; ++i){
		a[i] /= g;
		for(int j=0; prime[j]*prime[j] <= a[i]; ++j){
			int p = prime[j];
			if(a[i]%p == 0)	cnt[p]++;
			while(a[i]%p == 0)	a[i] /= p;
		}
		if(a[i] != 1)	cnt[a[i]]++;
	}
	// puts("*");
	int ans = n;
	for(int i=2; i<MAXN; ++i){
		// printf("%d\n", cnt[i]);
		ans = min(ans, n-cnt[i]);
	}
	if(ans == n)	ans = -1;
	printf("%d\n", ans);
	return 0;
}