이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define FORD(i, a, b) for (int i = (a); i >= (b); --i)
#define REP(i, a) for (int i = 0; i < (a); ++i)
#define DEBUG(x) { cerr << #x << '=' << x << endl; }
#define Arr(a, l, r) { cerr << #a << " = {"; FOR(_, l, r) cerr << ' ' << a[_]; cerr << "}\n"; }
#define N 1001000
#define pp pair<int, int>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define taskname ""
#define bit(S, i) (((S) >> (i)) & 1)
using namespace std;
double a[N], b[N];
int main() {
#ifdef NERO
freopen("test.inp","r",stdin);
freopen("test.out","w",stdout);
double stime = clock();
#else
//freopen(taskname".inp","r",stdin);
//freopen(taskname".out","w",stdout);
#endif //NERO
IO;
int n;
cin >> n;
FOR(i, 1, n) cin >> a[i] >> b[i];
sort(a + 1, a + n + 1, greater<double>());
sort(b + 1, b + n + 1, greater<double>());
double ans = 0, sa = 0, sb = 0;
int na = 0, nb = 0;
FOR(buy, 1, n * 2) {
if ((sa < sb && na < n) || sb == n) sa += a[++na];
else sb += b[++nb];
ans = max(ans, min(sa, sb) - 1.0 * buy);
}
cout << fixed << setprecision(4) << ans;
#ifdef NERO
double etime = clock();
cerr << "Execution time: " << (etime - stime) / CLOCKS_PER_SEC * 1000 << " ms.\n";
#endif // NERO
}
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