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QOJ

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
#127246	#6632. Minimize Median	Ethan_xu^#	WA	82ms	3548kb	C++20	2.1kb	2023-07-19 14:40:05	2023-07-19 14:40:09

Judging History

你现在查看的是最新测评结果

[2023-08-10 23:21:45]
System Update: QOJ starts to keep a history of the judgings of all the submissions.

[2023-07-19 14:40:09]
评测

测评结果：WA
用时：82ms
内存：3548kb

查看

[2023-07-19 14:40:05]
提交

answer

#include<bits/stdc++.h>
using namespace std ;

int main()
{
    ios::sync_with_stdio(false) ;
    cin.tie(0) ;


    int T ;
    cin >> T ;
    while(T --)
    {
        int n , m , k ;
        cin >> n >> m >> k ;
        vector<int> a(n + 1 , 0) ;
        for(int i = 1 ; i <= n ; i ++)  cin >> a[i] ;
        vector<int> cost(m + 2 , 0) ;
        for(int i = 1 ; i <= m ; i ++)  cin >> cost[i] ;
        cost[m + 1] = k + 1 ;
        for(int i = m - 1 ; i >= 1 ; i --)  cost[i] = min(cost[i] , cost[i + 1]) ;
        for(int i = 2 ; i <= m ; i ++)
        {
            int pp = min(i - 1 , (int)sqrt(i) + 2) ;
            for(int j = 1 ; j <= pp ; j ++)
                cost[i] = min(cost[i] , cost[j] + cost[(i + j - 1) / j]) ;
        }
        for(int i = m - 1 ; i >= 1 ; i --)  cost[i] = min(cost[i] , cost[i + 1]) ;
        int ppp = k + 1 ;
        for(int i = 2 ; i <= m - 1 ; i ++)  ppp = min(ppp , cost[i] + cost[(m + i - 1) / i]) ;
        cost[m] = min(cost[m] , ppp) ;
        for(int i = m - 1 ; i >= 1 ; i --)  cost[i] = min(cost[i] , cost[i + 1]) ;
        // cost[i] 表示使用了若干个炸弹，形成的伤害至少是i的最小代价和
        
        int l = 0 , r = m ;
        int res = r ;
        auto ok = [&](int x)
        {
            int p = n / 2 ;
            vector<int> v ;
            for(int i = 1 ; i <= n ; i ++)
                if(a[i] > x)
                    v.push_back(a[i]) ;
            if(v.size() <= p)  return true ;
            sort(v.begin() , v.end()) ;
            int cur = v.size() ;
            int t = k ;
            for(auto u : v)
            {
                int z = (u + 1 + x) / (x + 1) ; // u / z <= x
                assert(z <= m + 1) ;
                if(cost[z] <= t)  t -= cost[z] , cur -= 1 ;
                else  break ;
            }
            return cur <= p ;
        } ;
        while(l <= r)
        {
            int mid = (l + r) / 2 ;
            if(ok(mid))  res = mid , r = mid - 1 ;
            else  l = mid + 1 ;
        }
        cout << res << '\n' ;
    }

    return 0 ;
}
/*



*/

源文件, 语言: C++20

詳細信息

Test #1:

score: 100

Accepted

time: 1ms

memory: 3548kb

input:

output:

2
2
1

result:

ok 3 number(s): "2 2 1"

Test #2:

score: -100

Wrong Answer

time: 82ms

memory: 3448kb

input:

100000
5 10 5
3 7 1 10 10
11 6 11 6 1 8 9 1 3 1
5 6 51
2 2 2 5 1
42 61 26 59 100 54
5 10 76
7 5 8 4 7
97 4 44 83 61 45 24 88 44 44
5 8 90
1 1 5 1 3
35 15 53 97 71 83 26 7
5 3 52
1 1 3 1 1
22 6 93
5 6 28
6 6 1 3 1
9 31 2 19 10 27
5 8 31
3 6 2 1 2
32 29 13 7 57 34 9 5
5 6 75
3 3 4 5 4
40 56 38 60 17 3...

output:

result:

wrong answer 34th numbers differ - expected: '0', found: '1'