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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
#250585	#7678. The Game	Cu_OH_2	WA	4ms	3792kb	C++14	2.8kb	2023-11-13 13:39:17	2023-11-13 13:39:18

Judging History

你现在查看的是最新测评结果

[2023-11-13 13:39:18]
评测

测评结果：WA
用时：4ms
内存：3792kb

查看

[2023-11-13 13:39:17]
提交

answer

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

int n, m;

int back(multiset<int>& ms) { return *ms.rbegin(); }

int front(multiset<int>& ms) { return *ms.begin(); }

void erase(multiset<int>& ms, int x) { ms.erase(ms.find(x)); }

void pop_back(multiset<int>& ms) { ms.erase(prev(ms.end())); }

void add_front(multiset<int>& ms)
{
    int x = *ms.begin();
    ms.erase(ms.begin());
    ms.insert(x + 1);
}

void pop_front(multiset<int>& ms) { ms.erase(ms.begin()); }

void solve()
{
    cin >> n >> m;
    multiset<int> ms1, ms2;
    for (int i = 1; i <= n; ++i)
    {
        int x;
        cin >> x;
        ms1.insert(x);
    }
    for (int i = 1; i <= m; ++i)
    {
        int x;
        cin >> x;
        ms2.insert(x);
    }
    multiset<int> ms;
    vector<int> rec;
    while (ms2.size())
    {
        if (back(ms1) > back(ms2))
        {
            cout << "-1\n";
            return;
        }
        for (int i = back(ms1); i < back(ms2); ++i)
        {
            rec.push_back(i);
            if (rec.size() > n - m)
            {
                cout << "-1\n";
                return;
            }
        }
        ms.insert(back(ms1));
        pop_back(ms1);
        pop_back(ms2);
    }
    sort(rec.begin(), rec.end());
    vector<int> ans;
    int p = 0;
    // for (auto e : ms1) cerr << e << ' ';
    // cerr << endl;
    // for (auto e : ms) cerr << e << ' ';
    // cerr << endl;

    while (ms1.size())
    {
        if (ms1.size() > rec.size() - p)
        {
            while (front(ms1) == front(ms) && ms1.count(front(ms1)) == 1)
            {
                if (p >= rec.size())
                {
                    cout << "-1\n";
                    return;
                }
                ans.push_back(rec[p]);
                erase(ms, rec[p]);
                ms.insert(rec[p] + 1);
                p++;
                pop_front(ms1);
            }
            ans.push_back(front(ms1));
            add_front(ms1);
            pop_front(ms1);
        }
        else
        {
            ans.push_back(rec[p]);
            erase(ms, rec[p]);
            ms.insert(rec[p] + 1);
            p++;
            if (front(ms1) > front(ms))
            {
                cout << "-1\n";
                return;
            }
            pop_front(ms1);
        }
        // for (auto e : ms1) cerr << e << ' ';
        // cerr << "| ";
        // for (auto e : ms) cerr << e << ' ';
        // cerr << endl;
    }
    cout << ans.size() << '\n';
    for (auto e : ans) cout << e << ' ';
    cout << '\n';
    return;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
}

源文件, 语言: C++14

详细

Test #1:

score: 100

Accepted

time: 0ms

memory: 3792kb

input:

6
5 3
1 2 2 3 3
2 3 4
4 2
1 2 2 4
2 4
5 2
2 3 3 4 4
5 5
6 1
1 1 1 1 1 1
4
4 2
1 1 1 2
2 2
4 1
1 1 1 1
2

output:

result:

ok ok (6 test cases)

Test #2:

score: 0

Accepted

time: 4ms

memory: 3604kb

input:

output:

-1
1
1 
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
1
2 
-1
-1
-1
1
1 
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
...

result:

ok ok (7056 test cases)

Test #3:

score: -100

Wrong Answer

time: 3ms

memory: 3600kb

input:

output:

2
1 2 
-1
2
1 2 
-1
-1
-1
-1
2
1 1 
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
2
1 2 
-1
2
2 3 
-1
-1
-1
2
1 1 
2
1 2 
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
2
1 2 
-1
2
3 4 
-1
-1
-1
2
1 1 
2
1 3 
-1
-1
-1
2
1 2 
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
...

result:

wrong answer Wrong answer, the final sequence does not equal to B (test case 1)