D. Multiple Testcases （简单数学）

题目链接：https://codeforces.com/contest/1342/problem/D

想法：

我们首先预处理出 >= i 的个数

然后我们每次取遍历每个 i 然后可以得到最小的分组的个数

利用最小的组数我们可以从 1 开始跳着选取就好了

具体的还是看代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)

#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <ctime>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>

#define ll long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-10;
const int maxn = 2e5 + 10;
const int MOD = 998244353;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

int a[maxn], b[maxn], sum[maxn];

int main() {
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum[a[i]]++;
    }
    for (int i = 1; i <= k; i++) {
        cin >> b[i];
    }
    for (int i = k - 1; i >= 0; i--)
        sum[i] = sum[i + 1] + sum[i];
    sort(a + 1, a + 1 + n);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        ans = max(ans, sum[a[i]] / b[a[i]] + bool(sum[a[i]] % b[a[i]]));
    }
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) {
        int cnt = 0, now = i;
        while (now <= n) {
            cnt++;
            now += ans;
        }
        cout << cnt << " ";
        now = i;
        while (now <= n) {
            cout << a[now] << " ";
            now += ans;
        }
        cout << endl;
    }
    return 0;
}