結果

問題 No.817 Coin donation
ユーザー 👑 hitonanodehitonanode
提出日時 2019-04-19 21:34:04
言語 C++14
(gcc 11.2.0 + boost 1.78.0)
結果
AC  
実行時間 26 ms / 2,000 ms
コード長 3,452 bytes
コンパイル時間 1,211 ms
使用メモリ 3,972 KB
最終ジャッジ日時 2022-11-22 06:28:45
合計ジャッジ時間 2,405 ms
ジャッジサーバーID
（参考情報） 		judge14 / judge12
このコードへのチャレンジ（β）

テストケース

テストケース表示
入力 結果 実行時間
使用メモリ
testcase_00 AC 2 ms
3,460 KB
testcase_01 AC 1 ms
3,584 KB
testcase_02 AC 1 ms
3,536 KB
testcase_03 AC 2 ms
3,552 KB
testcase_04 AC 1 ms
3,592 KB
testcase_05 AC 1 ms
3,572 KB
testcase_06 AC 5 ms
3,504 KB
testcase_07 AC 6 ms
3,524 KB
testcase_08 AC 18 ms
3,576 KB
testcase_09 AC 6 ms
3,524 KB
testcase_10 AC 26 ms
3,772 KB
testcase_11 AC 24 ms
3,972 KB
testcase_12 AC 25 ms
3,884 KB
testcase_13 AC 1 ms
3,460 KB
testcase_14 AC 1 ms
3,516 KB
testcase_15 AC 1 ms
3,532 KB
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 # 

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))



int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> A(N), B(N);
    REP(i, N) cin >> A[i] >> B[i];

    lint l = 1, r = 1e9 + 2;
    while (r - l > 1)
    {
        lint c = (l + r) / 2;
        lint cou = 0;
        REP(i, N)
        {
            if (B[i] > c) cou += max(c - A[i] + 1, 0LL);
            else cou += B[i] - A[i] + 1;
        }
        (cou < K ? l : r) = c;
    }
    cout << r << endl;
}
0