結果
問題 | No.2374 ASKT Subsequences |
ユーザー | au7777 |
提出日時 | 2023-07-09 15:53:57 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,354 bytes |
コンパイル時間 | 4,381 ms |
コンパイル使用メモリ | 241,212 KB |
実行使用メモリ | 19,240 KB |
最終ジャッジ日時 | 2024-07-23 12:59:44 |
合計ジャッジ時間 | 5,846 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,816 KB |
testcase_01 | AC | 2 ms
6,940 KB |
testcase_02 | AC | 2 ms
6,940 KB |
testcase_03 | AC | 2 ms
6,940 KB |
testcase_04 | AC | 2 ms
6,940 KB |
testcase_05 | AC | 2 ms
6,944 KB |
testcase_06 | AC | 2 ms
6,940 KB |
testcase_07 | AC | 2 ms
6,940 KB |
testcase_08 | AC | 3 ms
6,940 KB |
testcase_09 | AC | 2 ms
6,940 KB |
testcase_10 | AC | 3 ms
6,944 KB |
testcase_11 | AC | 4 ms
6,940 KB |
testcase_12 | AC | 3 ms
6,940 KB |
testcase_13 | AC | 3 ms
6,944 KB |
testcase_14 | AC | 4 ms
6,940 KB |
testcase_15 | AC | 6 ms
7,436 KB |
testcase_16 | AC | 5 ms
6,944 KB |
testcase_17 | AC | 7 ms
8,484 KB |
testcase_18 | AC | 11 ms
11,140 KB |
testcase_19 | AC | 10 ms
10,540 KB |
testcase_20 | AC | 16 ms
13,968 KB |
testcase_21 | AC | 20 ms
15,188 KB |
testcase_22 | AC | 15 ms
12,888 KB |
testcase_23 | AC | 11 ms
11,356 KB |
testcase_24 | AC | 10 ms
11,312 KB |
testcase_25 | WA | - |
testcase_26 | AC | 31 ms
19,156 KB |
testcase_27 | WA | - |
testcase_28 | AC | 19 ms
19,084 KB |
testcase_29 | AC | 14 ms
19,240 KB |
testcase_30 | AC | 14 ms
19,100 KB |
ソースコード
#include <bits/stdc++.h> #include <atcoder/all> typedef long long int ll; using namespace std; typedef pair<ll, ll> P; using namespace atcoder; template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>; #define USE998244353 #ifdef USE998244353 const ll MOD = 998244353; // const double PI = 3.141592653589; using mint = modint998244353; #else const ll MOD = 1000000007; using mint = modint1000000007; #endif const int MAX = 20000001; long long fac[MAX], finv[MAX], inv[MAX]; void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll gcd(ll x, ll y) { if (y == 0) return x; else if (y > x) { return gcd (y, x); } else return gcd(x % y, y); } ll lcm(ll x, ll y) { return x / gcd(x, y) * y; } ll my_sqrt(ll x) { // ll m = 0; ll M = 3000000001; while (M - m > 1) { ll now = (M + m) / 2; if (now * now <= x) { m = now; } else { M = now; } } return m; } ll keta(ll num, ll arity) { ll ret = 0; while (num) { num /= arity; ret++; } return ret; } ll ceil(ll n, ll m) { // n > 0, m > 0 ll ret = n / m; if (n % m) ret++; return ret; } ll pow_ll(ll x, ll n) { if (n == 0) return 1; if (n % 2) { return pow_ll(x, n - 1) * x; } else { ll tmp = pow_ll(x, n / 2); return tmp * tmp; } } vector<ll> compress(vector<ll>& v) { // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] vector<ll> u = v; sort(u.begin(), u.end()); u.erase(unique(u.begin(),u.end()),u.end()); map<ll, ll> mp; for (int i = 0; i < u.size(); i++) { mp[u[i]] = i; } for (int i = 0; i < v.size(); i++) { v[i] = mp[v[i]]; } return v; } vector<ll> Eratosthenes( const ll N ) { vector<bool> is_prime( N + 1 ); for( ll i = 0; i <= N; i++ ) { is_prime[ i ] = true; } vector<ll> P; for( ll i = 2; i <= N; i++ ) { if( is_prime[ i ] ) { for( ll j = 2 * i; j <= N; j += i ) { is_prime[ j ] = false; } P.emplace_back( i ); } } return P; } int main() { int n; cin >> n; int a[n + 1]; int cum[n + 1][2001]; // i番目までにjが何個あるか for (int i = 0; i <= n; i++) { for (int j = 0; j <= 2000; j++) cum[i][j] = 0; } for (int i = 1; i <= n; i++) cin >> a[i]; for (int i = 1; i <= n; i++) { for (int j = 0; j <= 2000; j++) { if (j == a[i]) cum[i][j] = cum[i - 1][j] + 1; else cum[i][j] = cum[i - 1][j]; } } int ans = 0; for (int i = 2; i <= n - 2; i++) { for (int j = i + 1; j <= n - 1; j++) { if (a[i] <= a[j]) continue; if ((a[j] < 10) || (a[i] + 1 > 2000)) continue; ans += cum[i - 1][a[j] - 10] * (cum[n][a[i] + 1] - cum[j][a[i] + 1]); } } cout << ans << endl; return 0; }