Codeforces Round #511 (Div. 2) C. Enlarge GCD

链接：https://codeforces.com/contest/1047/problem/C

题意：给你n个数，删掉尽可能少的数，使公共gcd更大

题解：思想还是采用暴力枚举的思想，因为要使gcd更大，从gcd+1开始枚举，找出原序列中所有当前是gcd的倍数的数，最后n减去就是答案

代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<vector>
#include<stack>
#include<string>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<iomanip>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int M = int(1e7)*2 + 5;
int a[M],b[M];
int gcd(int a, int b)
{
    return  b ? gcd(b, a % b) : a;
}
signed main()
{
    int n; cin >> n;
    int g;
    int maxn = -1;
    for (int i = 0; i < n; i++){
        int x; cin >> x;
        a[x]++;
        if (i == 0)
            g = x;
        else
            g = gcd(g, x);
        maxn = max(maxn, x);
    }
    int ans = 0;
    for (int i = g + 1; i <= maxn; i++){
        if (!b[i]){
            int cnt = 0;
            for (int j = i; j <= maxn; j += i) {
                b[j] = 1;
                cnt += a[j];
            }
            ans = max(ans, cnt);
        }
    }
    if (ans == 0) cout << -1 << endl;
    else cout << n - ans << endl;
    return 0;
}