Непростые числа

#ifndef OLD
        #include <map>
        #include <set>
        #include <new>
        #include <stack>
        #include <deque>
        #include <queue>
        #include <cmath>
        #include <ctime>
        #include <locale>
        #include <memory>
        #include <vector>
        #include <cstdio>
        #include <limits>
        #include <cassert>
        #include <ctype.h>
        #include <climits>
        #include <cstdarg>
        #include <utility>
        #include <cstring>
        #include <iomanip>
        #include <numeric>
        #include <cassert>
        #include <sstream>
        #include <fstream>
        #include <complex>
        #include <iterator>
        #include <iostream>
        #include <typeinfo>
        #include <valarray>
        #include <algorithm>
        #include <functional>
#else
        #include <bits/stdc++.h>
#endif

#define f first
#define s second
#define re return
#define mp make_pair
#define pb push_back
#define sz(s) (int) s.size ()
#define all(s) s.begin (), s.end ()

#define name ""

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;

const int N = 100009;
const int inf = (int) 2e9;
const int mod = (int) 1e9 + 7;
const double eps = 1e-10;

int a[N];
int p[N];
int pr[N];
int dp[N][N];

int main () {
        int n;
        scanf ("%d", &n);
        for (int i = 0; i < n; i++) {
                scanf ("%d", a + i);
        }
        int m = 0;
        for (int i = 2; i <= 2000; i++) {
                if (!pr[i]) {
                        p[m++] = i;
                        for (int j = i * i; j <= 2000; j += i) {
                                pr[j] = 1;
                        }
                }
        }
        for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                        dp[i][j] = inf;
                }
        }
        sort (a, a + n);
        int ans = inf;
        for (int i = 0; i < m; i++) {
                dp[0][i] = abs (a[0] - p[i]);
                if (n == 1) {
                        ans = min (ans, dp[0][i]);
                }
        }
        for (int i = 1; i < n; i++) {
                for (int j = i - 1; j < m; j++) {
                        for (int k = j + 1; k < m; k++) {
                                dp[i][k] = min (dp[i][k], dp[i - 1][j] + abs (a[i] - p[k]));
                                if (i == n - 1) {
                                        ans = min (ans, dp[i][k]);
                                }
                        }
                }
        }
        printf ("%d\n", ans);
        return 0;
}