Codeforces Round #734 (Div. 3)

B2
签到

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;

int n, m;
map<int, int> mp;
vector<int> v[N];
int a[N];
int ans[N];

void solve()
{
    mp.clear();
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        mp[a[i]]++;
        v[a[i]].push_back(i);
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++)
        cnt += (mp[i] >= m ? m : mp[i]);
    cnt = cnt / m;
    int p = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j < v[i].size(); j++)
        {
            if(j < m && cnt)
            {
                ans[v[i][j]] = p++;
                if(p == m+1)
                {
                    cnt--;
                    p = 1;
                }
            }
            else
                ans[v[i][j]] = 0;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        cout << ans[i] << ' ';
        v[i].clear();
    }
    cout << endl;
}

int main()
{
    int T;
    for (cin >> T; T--;)
    {
        solve();
    }
        return 0;
}

D1
解题思路：分奇偶讨论，如果n,m都为偶数，则只需满足k为偶数即可，如果n为奇数，需要把奇数行先填为偶数行，然后k为偶数即可。如果m为奇数，需要满足k<=(m/2)*n，且k为偶数。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;


void solve()
{
    int n, m, k;
    cin >> n >> m >> k;
    if(n&1)
    {
        if((k-m/2)>=0 && (k-m/2)%2==0)
            puts("YES");
        else
            puts("NO");
    }
    else if(m&1)
    {
        if(k%2==0&&k<=(m/2)*n)
            puts("YES");
        else
            puts("NO");
    }
    else 
    {
        if(k&1)
            puts("NO");
        else
            puts("YES");
    }
}

int main()
{
    int T;
    for (cin >> T; T--;)
    {
        solve();
    }
        return 0;
}

D2
E
解题思路：数据为2e3，容易想到用dp，dp{i，j}表示，到当前i个元素，删掉了j个元素的最大匹配程度。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;
int dp[N][N];
int a[N];

void solve()
{
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= i; j++)
        {
            if(j)
                dp[i][j] = max(dp[i][j], dp[i-1][j-1]);
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + (a[i] == (i - j)));
        }
    for (int i = 1; i <= n; i++)
    {
        if(dp[n][i] >= k)
        {
            cout << i << endl;
            return;
        }
    }
    cout << -1 << endl;
}

int main()
{
    int T;
    for (cin >> T; T--;)
    {
        solve();
    }
        return 0;
}