Codeforces 1342 D. Multiple Testcases

题目入口
题意：给你n个数的数列m和k个数的数列c，要求将m序列分成经量少的组，每组必须满足不超过c[i]个数大于等于i的条件。

思路：

• 用sum[i]记录大于等于i的个数，那么ans = max(ans，sum[i] / c[i])，其中sum[i] / c[i]上取整。
• 再将原序列m升序排列，直接将其i % ans 对号入座进入答案组中。

代码实现：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <iostream>
#include <sstream>
#include <string>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#define int long long
#define lowbit(x) (x &(-x))
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const int  INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double EXP = 1e-8;
const ll   MOD = 1e9 + 7;
const int  N = 2e5 + 5;

int n, k, num;
int m[N], c[N], sum[N];
vector<int> ans[N];

signed main()
{
    IOS;
    cin >> n >> k;
    for(int i = 0; i < n; i ++){
        cin >> m[i];
        sum[m[i]] ++;
    }
    sort(m, m + n);
    for(int i = 1; i <= k; i ++)
        cin >> c[i];

    for(int i = k; i; i --){
        sum[i] += sum[i + 1];
        double tmp = sum[i] * 1.0 / c[i];
        if(tmp > (int)tmp) tmp ++;
        num = max(num, (int)tmp);
    }

    for(int i = 0; i < n; i ++)
        ans[i % num].push_back(m[i]);

    cout << num << endl;
    for(int i = 0; i < num; i ++){
        cout << ans[i].size();
        for(auto it : ans[i]) cout << " " << it;
        cout << endl;
    }

    return 0;
}

优先队列不熟悉，找了好久的规律，嗐~
学长说这个题还可以用背包做，%%%.