結果

問題 No.1557 Binary Variable
ユーザー rniya
提出日時 2021-06-25 22:26:09
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 4,575 bytes
コンパイル時間 2,170 ms
コンパイル使用メモリ 197,164 KB
最終ジャッジ日時 2025-01-22 12:23:48
ジャッジサーバーID
(参考情報)
judge5 / judge3
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 * 24 RE * 7 MLE * 3
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#pragma region Macros
typedef long long ll;
typedef __int128_t i128;
typedef unsigned int uint;
typedef unsigned long long ull;
#define ALL(x) (x).begin(), (x).end()
template <typename T> istream& operator>>(istream& is, vector<T>& v) {
for (T& x : v) is >> x;
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 == (int)v.size() ? "" : " ");
}
return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const pair<T, U>& p) {
os << '(' << p.first << ',' << p.second << ')';
return os;
}
template <typename T, typename U, typename V> ostream& operator<<(ostream& os, const tuple<T, U, V>& t) {
os << '(' << get<0>(t) << ',' << get<1>(t) << ',' << get<2>(t) << ')';
return os;
}
template <typename T, typename U, typename V, typename W> ostream& operator<<(ostream& os, const tuple<T, U, V, W>& t) {
os << '(' << get<0>(t) << ',' << get<1>(t) << ',' << get<2>(t) << ',' << get<3>(t) << ')';
return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const map<T, U>& m) {
os << '{';
for (auto itr = m.begin(); itr != m.end();) {
os << '(' << itr->first << ',' << itr->second << ')';
if (++itr != m.end()) os << ',';
}
os << '}';
return os;
}
template <typename T, typename U> ostream& operator<<(ostream& os, const unordered_map<T, U>& m) {
os << '{';
for (auto itr = m.begin(); itr != m.end();) {
os << '(' << itr->first << ',' << itr->second << ')';
if (++itr != m.end()) os << ',';
}
os << '}';
return os;
}
template <typename T> ostream& operator<<(ostream& os, const set<T>& s) {
os << '{';
for (auto itr = s.begin(); itr != s.end();) {
os << *itr;
if (++itr != s.end()) os << ',';
}
os << '}';
return os;
}
template <typename T> ostream& operator<<(ostream& os, const multiset<T>& s) {
os << '{';
for (auto itr = s.begin(); itr != s.end();) {
os << *itr;
if (++itr != s.end()) os << ',';
}
os << '}';
return os;
}
template <typename T> ostream& operator<<(ostream& os, const unordered_set<T>& s) {
os << '{';
for (auto itr = s.begin(); itr != s.end();) {
os << *itr;
if (++itr != s.end()) os << ',';
}
os << '}';
return os;
}
template <typename T> ostream& operator<<(ostream& os, const deque<T>& v) {
for (int i = 0; i < (int)v.size(); i++) {
os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
}
return os;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail> void debug_out(Head&& head, Tail&&... tail) {
cerr << head;
if (sizeof...(Tail) > 0) cerr << ", ";
debug_out(move(tail)...);
}
#ifdef LOCAL
#define debug(...) \
cerr << " "; \
cerr << #__VA_ARGS__ << " :[" << __LINE__ << ":" << __FUNCTION__ << "]" << '\n'; \
cerr << " "; \
debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
template <typename T> T gcd(T x, T y) { return y != 0 ? gcd(y, x % y) : x; }
template <typename T> T lcm(T x, T y) { return x / gcd(x, y) * y; }
template <class T1, class T2> inline bool chmin(T1& a, T2 b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <class T1, class T2> inline bool chmax(T1& a, T2 b) {
if (a < b) {
a = b;
return true;
}
return false;
}
#pragma endregion
const int INF = 1e9;
const long long IINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const char dir[4] = {'D', 'R', 'U', 'L'};
const long long MOD = 1000000007;
// const long long MOD = 998244353;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N, M;
cin >> N >> M;
vector<vector<int>> check(N + 1);
for (int i = 0; i < M; i++) {
int L, R;
cin >> L >> R;
check[R].emplace_back(--L);
}
vector<int> sum(N + 1, 0);
for (int r = 1; r <= N; r++) {
sum[r] = sum[r - 1];
bool ok = true;
for (int& l : check[r]) {
if (sum[r] - sum[l] == 0) {
ok = false;
}
}
if (!ok) sum[r]++;
}
int ans = N - sum.back();
cout << ans << '\n';
return 0;
}
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