이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <set>
using namespace std;
const int maxN = 3e3 + 5;
long long sarr[maxN], unreachable[maxN];
set<pair<long long, long long> > s;
long long A, B, C, T;
int main(){
int n, m, k;
cin >> n >> m >> k;
cin >> A >> B >> C >> T;
sarr[0] = 0;
for (int i = 1; i <= m; i++) cin >> sarr[i], sarr[i]--;
sarr[m+1] = n;
int ans = 0;
for (int i = 0; i <= m; i++){
if (sarr[i] == sarr[i+1]) continue;
long long cons = sarr[i] * B;
unreachable[i] = (T-cons+A)/A;
unreachable[i] = min(unreachable[i], sarr[i+1] - sarr[i]);
unreachable[i] = max(unreachable[i], 0ll);
ans += unreachable[i];
}
for (int i = 0; i <= m; i++){
if (sarr[i] == sarr[i+1]) continue;
long long cons = sarr[i]*B + unreachable[i]*C;
long long next = (T-cons+A)/A + unreachable[i];
next = min(next, sarr[i+1] - sarr[i]);
next = max(next, unreachable[i]);
s.insert(make_pair(next-unreachable[i], i));
unreachable[i] = next;
}
int count = 0;
while(count < k-m){
pair<long long, long long> best = *s.rbegin(); s.erase(--s.end());
ans += best.first;
int i = best.second;
long long cons = sarr[i]*B + unreachable[i]*C;
long long next = (T-cons+A)/A + unreachable[i];
next = min(next, sarr[i+1] - sarr[i]);
next = max(next, unreachable[i]);
s.insert(make_pair(next-unreachable[i], i));
unreachable[i] = next;
count++;
}
cout << ans-1 << endl;
}
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