結果

問題 No.1095 Smallest Kadomatsu Subsequence
ユーザー umnc
提出日時 2022-08-22 23:09:39
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 1,941 bytes
コンパイル時間 2,416 ms
コンパイル使用メモリ 197,092 KB
最終ジャッジ日時 2025-01-31 02:58:41
ジャッジサーバーID
(参考情報)
judge2 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 29 WA * 1
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
// #include <atcoder/all>
// using namespace atcoder;
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= n; ++i)
#define reps(i, s, e) for (ll i = s; i <= e; ++i)
#define rrep(i, n) for (ll i = n - 1; 0 <= i; --i)
#define all(v) v.begin(), v.end()
#define endl "\n"
template <class T>
bool chmax(T& a, const T& b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
if (b < a) {
a = b;
return 1;
}
return 0;
}
using ll = long long;
using ld = long double;
using cp = complex<ld>;
using pa = pair<ll, ll>;
using tup = tuple<ll, ll, ll>;
using vp = vector<pair<ll, ll> >;
using vtup = vector<tuple<ll, ll, ll> >;
using st = string;
using vs = vector<string>;
using vc = vector<char>;
using vvi = vector<vector<ll> >;
using vvc = vector<vector<char> >;
using vi = vector<ll>;
const ll MOD1 = 1000000007;
const ll MOD2 = 998244353;
const ll INF = 1000000000;
void init() {
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
cout << fixed << setprecision(15);
}
int main() {
init();
ll n;
cin >> n;
vi a(n), lmn(n + 1), rmn(n + 1), lmx(n + 1), rmx(n + 1);
lmn[0] = rmn[0] = INF;
rep(i, n) cin >> a[i];
rep(i, n) {
lmn[i + 1] = min(lmn[i], a[i]);
lmx[i + 1] = max(lmx[i], a[i]);
rmn[i + 1] = min(rmn[i], a[n - i - 1]);
rmx[i + 1] = max(rmx[i], a[n - i - 1]);
}
ll ans = -1;
reps(i, 1, n - 1) {
if (lmn[i] < a[i] && rmn[n - i - 1] < a[i]) {
if (ans == -1) {
ans = a[i] + lmn[i] + rmn[n - i - 1];
} else {
ans = min(ans, a[i] + lmn[i] + rmn[n - i - 1]);
}
}
if (lmn[i] > a[i] && rmn[n - i - 1] > a[i]) {
if (ans == -1) {
ans = a[i] + lmn[i] + rmn[n - i - 1];
} else {
ans = min(ans, a[i] + lmn[i] + rmn[n - i - 1]);
}
}
}
cout << ans << endl;
}
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