Codeforces Round #432 (Div. 2) D. Arpa and a list of numbers（暴力）

枚举质数，判断是否超过临界值。临界值就是将不是因子中不含这个素数的数的个数乘以x和y的较小值，是否小于当前最小值。

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>

#define Max 1000001
#define MAXN 500005
#define MAXNUM 1000005
#define MOD 100000000
#define rin freopen("in.txt","r",stdin)
#define rout freopen("1.out","w",stdout)
#define Del(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const LL LINF = 0x3f3f3f3f3f3f3f3fLL;

int a[MAXN];
int sum[MAXNUM];
int vis[MAXNUM];
int n, x, y;
int main() {
    //rin;
    scanf("%d%d%d", &n, &x, &y);
    Del(sum, 0);
    Del(vis, 0);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        sum[a[i]]++;
    }
    LL cnt = LINF;

    for (int i = 2; i <= 1000000; i++) {
        if (!vis[i]) {
            LL ans = sum[i];
            for (int j = 2 * i; j <= 1000000; j += i)
                vis[j] = 1, ans += sum[j];
            if ((n - ans) * min(x, y) < cnt) {
                LL tans = 0;
                for (int j = 0; j < n; j++) {
                    if (a[j] % i) {
                        LL t = a[j] % i;
                        tans += min(1ll * x, 1ll * y * (i - t));
                    }
                }
                cnt = min(cnt, tans);
            }
        }
    }
    printf("%lld",cnt);
    return 0;
}