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# solution_uit_55_180.cpp

Oct 1st, 2022
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1. #include <bits/stdc++.h>
2. using namespace std;
3.
4. typedef long long ll;
5.
6. set<int> prime;
7. ll mem[17][9 * 17 + 1][2];
8.
9. bool isPrime(int n) {
10.     if (n < 2) return false;
11.     for (int i = 2; i * i <= n; i++) {
12.         if (n % i == 0) return false;
13.     }
14.     return true;
15. }
16.
17. ll dp(string &s, int pos, int sum, bool lwr) {
18.     if (pos == (int) s.size()) return prime.count(sum);
19.     if (mem[pos][sum][lwr] != -1) return mem[pos][sum][lwr];
20.     ll ret = 0;
21.     for (int digit = 0; digit <= (lwr ? s[pos] - '0' : 9); digit++) {
22.         ret += dp(s, pos + 1, sum + digit, (lwr && digit == (s[pos] - '0')));
23.     }
24.     // cout << pos << ' ' << sum << ' ' << lwr << ' ' << ret << '\n';
25.     return mem[pos][sum][lwr] = ret;
26. }
27.
28. int main() {
29.     cin.tie(0)->sync_with_stdio(0);
30.     for (int i = 2; i <= 9 * 17; i++) {
31.         if (isPrime(i)) prime.insert(i);
32.     }
33.     ll k;
34.     cin >> k;
35.     ll l = 2, r = 1e17, ans = -1;
36.     while (l <= r) {
37.         memset(mem, -1, sizeof(mem));
38.         ll m = (l + r) / 2;
39.         string s = to_string(m);
40.         if (dp(s, 0, 0, 1) >= k) {
41.             ans = m;
42.             r = m - 1;
43.         }
44.         else l = m + 1;
45.     }
46.     cout << ans << '\n';
47. }
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