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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#718532#5417. Chat ProgramRighttRE 1ms5680kbC++141.4kb2024-11-06 20:46:042024-11-06 20:46:07

Judging History

你现在查看的是最新测评结果

  • [2024-11-06 20:46:07]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:5680kb
  • [2024-11-06 20:46:04]
  • 提交

answer

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

using namespace std;
using ll = long long;

const int N = 2e5 + 10;

ll n, k, m, c, d;
ll a[N], p[N];

bool check(ll x){
    //难写!
    //预处理<x的时刻
    for (int i = 0; i < n; i ++ ) p[i] = 0;
    int res = 0;
    for (int i = 0; i < n; i ++ ){
        res += a[i] >= x;
    }
    //前m
    for (int i = 0; i < m; i ++ ){
        if (a[i] >= x || a[i] + c + d * i < x) continue;
        res ++;
        //贡献-1
        int pre ;//不行位置
        if (d == 0 || a[i] + c >= x){
            pre = i + 1;
        }else{
            //至少加y次
            int y = (x - c - a[i] + d - 1) / 1;
            pre = i - y;
            pre ++;
        }
        p[pre] ++;
    }
    if (res >= k) return true;
    for (int i = 1; i + m - 1 < n; i ++ ){
        if (a[i+m-1] < x && a[i+m-1] + c + d * (m - 1) >= x) res ++;
        res -= p[i];
        if (res >= k) return true;
    }
    return false;
}

void solve(){
    cin >> n >> k >> m >> c >> d;
    for (int i = 0; i < n; i ++ ) cin >> a[i];

    ll l = 0, r = 1e9 * (2e5 + 10);
    while (l < r){
        ll mid = (l + r + 1) / 2;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l;
}

signed main(){
    IOS;
    int t = 1;
//		cin >> t;
    while (t -- ){
        solve();
    }
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 5680kb

input:

6 4 3 1 2
1 1 4 5 1 4

output:

4

result:

ok 1 number(s): "4"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3556kb

input:

7 3 2 4 0
1 9 1 9 8 1 0

output:

9

result:

ok 1 number(s): "9"

Test #3:

score: 0
Accepted
time: 0ms
memory: 5624kb

input:

8 3 5 0 0
2 0 2 2 1 2 1 8

output:

2

result:

ok 1 number(s): "2"

Test #4:

score: -100
Runtime Error

input:

200000 200000 100000 0 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...

output:


result: