結果
問題 |
No.3244 Range Multiple of 8 Query
|
ユーザー |
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提出日時 | 2025-08-07 01:29:10 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 1,491 ms / 5,000 ms |
コード長 | 2,773 bytes |
コンパイル時間 | 1,083 ms |
コンパイル使用メモリ | 89,060 KB |
実行使用メモリ | 54,360 KB |
最終ジャッジ日時 | 2025-08-07 01:29:43 |
合計ジャッジ時間 | 31,866 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 40 |
ソースコード
#include <algorithm> #include <iostream> #include <string> #include <vector> using namespace std; using ll = long long; ll inv(ll x1, ll x2, ll x3) { ll ret = 0; if (x1 > x2) { ret++; } if (x1 > x3) { ret++; } if (x2 > x3) { ret++; } return ret; } int main() { ll N, Q; cin >> N >> Q; string S; cin >> S; vector<ll> l(Q); vector<ll> r(Q); for (ll i = 0; i < Q; i++) { cin >> l[i] >> r[i]; } vector<vector<vector<int>>> prev_idx( 3, vector<vector<int>>(10, vector<int>(N + 1, -1))); for (ll i = 0; i < N; i++) { ll val = (ll)(S[i] - '0'); for (ll j = 1; j < 10; j++) { if (val == j) { prev_idx[0][j][i + 1] = i + 1; for (ll k = 0; k < 2; k++) { prev_idx[k + 1][j][i + 1] = prev_idx[k][j][i]; } } else { for (ll k = 0; k < 3; k++) { prev_idx[k][j][i + 1] = prev_idx[k][j][i]; } } } } vector<ll> mul_8(0); for (ll i = 0; i < 1000; i += 8) { string sr = to_string(1000 + i); bool inc_0 = false; for (ll j = 0; j < 4; j++) { if (sr[j] == '0') { inc_0 = true; } } if (!inc_0) { mul_8.push_back(i); } } ll INF = 100000000000023; vector<ll> vec_cnt(10, 0); for (ll i = 0; i < Q; i++) { if (l[i] == r[i]) { if (S[l[i] - 1] == '8') { cout << 0 << endl; } else { cout << -1 << endl; } continue; } else if (l[i] + 1 == r[i]) { string fs = S.substr(l[i] - 1, 2); string rs = fs; reverse(rs.begin(), rs.end()); if (stoll(fs) % 8 == 0) { cout << 0 << endl; } else if (stoll(rs) % 8 == 0) { cout << 1 << endl; } else { cout << -1 << endl; } continue; } ll ans = INF; for (ll x : mul_8) { ll t1 = x % 10; ll t2 = (x / 10) % 10; ll t3 = x / 100; ll id1 = prev_idx[vec_cnt[t1]][t1][r[i]]; vec_cnt[t1]++; ll id2 = prev_idx[vec_cnt[t2]][t2][r[i]]; vec_cnt[t2]++; ll id3 = prev_idx[vec_cnt[t3]][t3][r[i]]; vec_cnt[t1]--; vec_cnt[t2]--; if (id1 < l[i] || id2 < l[i] || id3 < l[i]) { continue; } ans = min(ans, 3 * r[i] - 3 - id1 - id2 - id3 + inv(id3, id2, id1)); } cout << (ans == INF ? -1 : ans) << endl; } }