結果

問題 No.1368 サイクルの中に眠る門松列
ユーザー Ricky_pon
提出日時 2021-01-29 21:45:57
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 41 ms / 2,000 ms
コード長 1,918 bytes
コンパイル時間 1,811 ms
コンパイル使用メモリ 201,324 KB
最終ジャッジ日時 2025-01-18 09:07:17
ジャッジサーバーID
(参考情報)
judge5 / judge4
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 * 1
other AC * 15
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘void solve()’:
main.cpp:67:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   67 |     scanf("%d", &n);
      |     ~~~~~^~~~~~~~~~
main.cpp:69:20: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   69 |     rep(i, n) scanf("%lld", &a[i]);
      |               ~~~~~^~~~~~~~~~~~~~~
main.cpp: In function ‘int main()’:
main.cpp:82:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   82 |     scanf("%d", &t);
      |     ~~~~~^~~~~~~~~~

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>
//#include <atcoder/all>
#define For(i, a, b) for (int(i) = (int)(a); (i) < (int)(b); ++(i))
#define rFor(i, a, b) for (int(i) = (int)(a)-1; (i) >= (int)(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;
template <class T>
bool chmax(T& a, const T& b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <class T>
bool chmin(T& a, const T& b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <class T>
T div_floor(T a, T b) {
if (b < 0) a *= -1, b *= -1;
return a >= 0 ? a / b : (a + 1) / b - 1;
}
template <class T>
T div_ceil(T a, T b) {
if (b < 0) a *= -1, b *= -1;
return a > 0 ? (a - 1) / b + 1 : a / b;
}
constexpr lint mod = 1000000007;
constexpr lint INF = mod * mod;
constexpr int MAX = 100010;
bool check(deque<lint>& a, int i) {
if (a[i] == a[i + 1] || a[i + 1] == a[i + 2] || a[i] == a[i + 2])
return false;
return *max_element(a.begin() + i, a.begin() + i + 3) == a[i + 1] ||
*min_element(a.begin() + i, a.begin() + i + 3) == a[i + 1];
}
lint solve(int n, deque<lint>& a) {
lint dp[n + 1];
memset(dp, 0, sizeof(dp));
rep(i, n) {
chmax(dp[i + 1], dp[i]);
if (i + 3 <= n && check(a, i)) {
chmax(dp[i + 3], dp[i] + a[i]);
}
}
return dp[n];
}
void solve() {
int n;
scanf("%d", &n);
deque<lint> a(n);
rep(i, n) scanf("%lld", &a[i]);
lint ans = 0;
rep(_, 3) {
chmax(ans, solve(n, a));
auto x = a.front();
a.pop_front();
a.push_back(x);
}
printf("%lld\n", ans);
}
int main() {
int t;
scanf("%d", &t);
rep(tt, t) solve();
}
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