結果
問題 | No.1095 Smallest Kadomatsu Subsequence |
ユーザー |
![]() |
提出日時 | 2020-07-12 22:29:37 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 116 ms / 2,000 ms |
コード長 | 3,539 bytes |
コンパイル時間 | 1,735 ms |
コンパイル使用メモリ | 177,048 KB |
実行使用メモリ | 19,012 KB |
最終ジャッジ日時 | 2024-11-06 05:46:01 |
合計ジャッジ時間 | 4,774 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 30 |
ソースコード
#include <bits/stdc++.h>using namespace std;//#define int long longtypedef long long ll;typedef unsigned long long ul;typedef unsigned int ui;const ll mod = 1000000007;const ll INF = mod * mod;const int INF_N = 1e+9;typedef pair<int, int> P;#define stop char nyaa;cin>>nyaa;#define rep(i,n) for(int i=0;i<n;i++)#define per(i,n) for(int i=n-1;i>=0;i--)#define Rep(i,sta,n) for(int i=sta;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define per1(i,n) for(int i=n;i>=1;i--)#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)#define all(v) (v).begin(),(v).end()typedef pair<ll, ll> LP;typedef long double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-12;const ld pi = acos(-1.0);//typedef vector<vector<ll>> mat;typedef vector<int> vec;//繰り返し二乗法ll mod_pow(ll a, ll n, ll m) {ll res = 1;while (n) {if (n & 1)res = res * a%m;a = a * a%m; n >>= 1;}return res;}struct modint {ll n;modint() :n(0) { ; }modint(ll m) :n(m) {if (n >= mod)n %= mod;else if (n < 0)n = (n%mod + mod) % mod;}operator int() { return n; }};bool operator==(modint a, modint b) { return a.n == b.n; }modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }modint operator+(modint a, modint b) { return a += b; }modint operator-(modint a, modint b) { return a -= b; }modint operator*(modint a, modint b) { return a *= b; }modint operator^(modint a, int n) {if (n == 0)return modint(1);modint res = (a*a) ^ (n / 2);if (n % 2)res = res * a;return res;}//逆元(Eucledean algorithm)ll inv(ll a, ll p) {return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);}modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }const int max_n = 1 << 18;modint fact[max_n], factinv[max_n];void init_f() {fact[0] = modint(1);for (int i = 0; i < max_n - 1; i++) {fact[i + 1] = fact[i] * modint(i + 1);}factinv[max_n - 1] = modint(1) / fact[max_n - 1];for (int i = max_n - 2; i >= 0; i--) {factinv[i] = factinv[i + 1] * modint(i + 1);}}modint comb(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[b] * factinv[a - b];}using mP = pair<modint, modint>;int dx[4] = { 0,1,0,-1 };int dy[4] = { 1,0,-1,0 };void solve() {int N; cin >> N;vec a(N);ll res = INF;int num = 0;rep(i, N){cin >> a[i];if(res > a[i]){res = a[i];num = i;}}vec mi(N), rmi(N);rep(i, N){if(i == 0) mi[i] = a[i];else {mi[i] = min(mi[i-1], a[i]);}}per(i, N){if(i == N-1) rmi[i] = a[i];else{rmi[i] = min(rmi[i+1], a[i]);}}ll res1 = INF;Rep(i, 1, N-1){if(mi[i-1] < a[i] && a[i] > rmi[i+1]) res1 = min(res1, (ll)mi[i-1]+a[i]+rmi[i+1]);}ll res2 = INF;set<int> s1, s2;if(num != 0 && num != N-1){rep(i, num) s1.insert(a[i]);Rep(i, num+1, N) s2.insert(a[i]);res2 = *s1.begin() + a[num] + *s2.begin();}ll ans = min(res1, res2);if(ans == INF) cout << -1 << endl;else cout << ans << endl;}signed main() {ios::sync_with_stdio(false);cin.tie(0);//cout << fixed << setprecision(10);//init_f();//init();//int t; cin >> t; rep(i, t)solve();solve();// stopreturn 0;}