結果
問題 | No.1077 Noelちゃんと星々4 |
ユーザー | hirono999 |
提出日時 | 2020-06-12 21:59:54 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
RE
|
実行時間 | - |
コード長 | 6,106 bytes |
コンパイル時間 | 2,404 ms |
コンパイル使用メモリ | 203,416 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-06-24 05:01:34 |
合計ジャッジ時間 | 4,187 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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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
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | RE | - |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | RE | - |
testcase_06 | WA | - |
testcase_07 | WA | - |
testcase_08 | WA | - |
testcase_09 | RE | - |
testcase_10 | RE | - |
testcase_11 | WA | - |
testcase_12 | WA | - |
testcase_13 | RE | - |
testcase_14 | WA | - |
testcase_15 | WA | - |
testcase_16 | WA | - |
testcase_17 | WA | - |
testcase_18 | WA | - |
testcase_19 | WA | - |
testcase_20 | AC | 2 ms
5,376 KB |
testcase_21 | WA | - |
ソースコード
#include <bits/stdc++.h> const long long INF = 1LL << 60; const long long MOD = 1000000007; const double PI = acos(-1.0); #define rep(i, n) for (ll i = 0; i < (n); ++i) #define rep1(i, n) for (ll i = 1; i <= (n); ++i) #define rrep(i, n) for (ll i = (n - 1); i >= 0; --i) #define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c))) #define ALL(obj) (obj).begin(), (obj).end() #define RALL(obj) (obj).rbegin(), (obj).rend() #define pb push_back #define to_s to_string #define len(v) (ll)v.size() #define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) #define print(x) cout << (x) << '\n' #define drop(x) cout << (x) << '\n', exit(0) #define debug(x) cout << #x << ": " << (x) << '\n' using namespace std; using ll = long long; typedef pair<ll, ll> P; typedef vector<ll> vec; typedef vector<vector<ll>> vec2; typedef vector<vector<vector<ll>>> vec3; template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; } template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; } inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); } inline ll devc(ll x, ll y) { return (x + y - 1) / y; } inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); } struct IoSetup { IoSetup() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(10); cerr << fixed << setprecision(10); } } iosetup; template< typename T1, typename T2 > ostream &operator << (ostream &os, const pair< T1, T2 > &p) { os << p.first << " " << p.second; return os; } template< typename T1, typename T2 > istream &operator >> (istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; } template< typename T1, typename T2, typename T3 > ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) { os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t); return os; } template< typename T1, typename T2, typename T3 > istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return is; } template< typename T > ostream &operator << (ostream &os, const vector< T > &v){ for (int i = 0; i < (int)v.size(); ++i) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } template< typename T > istream &operator >> (istream &is, vector< T > &v){ for(T &in : v) is >> in; return is; } /*--------------------------------- Tools ------------------------------------------*/ template< typename T > vector<T> cumsum(const vector<T> &X){ vector<T> res(X.size() + 1, 0); for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i]; return res; } template< typename S, typename T, typename F> pair<T, T> bisearch(S left, T right, F f) { while(abs(right - left) > 1){ T mid = (right + left) / 2; if(f(mid)) right = mid; else left = mid; } return {left, right}; } template< typename S, typename T, typename F> double trisearch(S left, T right, F f, int maxLoop = 90){ double low = left, high = right; while(maxLoop--){ double mid_left = high / 3 + low * 2 / 3; double mid_right = high * 2 / 3 + low / 3; if(f(mid_left) >= f(mid_right)) low = mid_left; else high = mid_right; } return (low + high) * 0.5; } template< typename F > ll ternarySearch(ll L, ll R, F f) { //[L, R) ll lo = L - 1, hi = R - 1; while (lo + 1 != hi) { ll mi = (lo + hi) / 2; if (f(mi) <= f(mi + 1)) hi = mi; else lo = mi; } return hi; } //Def of Monoid //Suppose that S is a set and ● is some binary opeartion S x S -> S //then S with ● is a monoid if it satisfies the following two: // Associativity(結合則) // For all a,b and c in S, the equation (a ● b) ● c = a ● (b ● c) holds // Identitiy element(単位元の存在) // There exisits an element e in S such that for every element a in S, // the equations e ● a = a ● e = a holds //Eample of Monoid //+, *, and, or, xor, min, max //Build O(N) //Query O(log N) //- query(a,b) : applay operation to the range [a, b) //- update(k,x) : change k-th element to x //- operator[k] : return k-th element template< typename Monoid > class SegmentTree{ private: using F = function<Monoid(Monoid, Monoid)>; long long sz; vector<Monoid> seg; const F f; const Monoid e; public: SegmentTree(long long n, const F f, const Monoid &e) : f(f), e(e){ sz = 1; while(sz < n) sz <<= 1; seg.assign(2 * sz, e); } void set(long long k, const Monoid &x){ seg[k + sz] = x; } void build() { for (long long k = sz - 1; k > 0; --k){ seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]); } } void update(long long k, const Monoid &x) { k += sz; seg[k] = x; while(k >>= 1) seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]); } Monoid query(long long a, long long b){ Monoid L = e, R = e; for (a += sz, b += sz; a < b; a >>= 1, b >>= 1){ if(a & 1) L = f(L, seg[a++]); if(b & 1) R = f(seg[--b],R); } return f(L, R); } Monoid operator[](const int &k) const { return seg[k + sz]; } }; /*------------------------------- Main Code Here -----------------------------------------*/ int main() { ll N; cin >> N; vec Y(N); cin >> Y; vec Z(N); rep(i, N) if(i >= 1) Z[i] = Y[i] - Y[i - 1]; ll ans = 0; rep(i, N){ if(i + 1 >= 0) if(Z[i + 1] < 0 and Z[i] > 0){ ans += min(Z[i], abs(Z[i + 1])); Z[i + 1] += min(Z[i], abs(Z[i + 1])); } if(Z[i] < 0){ ans += abs(Z[i]); if(i + 1 >= 0) Z[i + 1] -= abs(Z[i]); } } print(ans); return 0; }