codeforces 1342 D 贪心+后缀和

翻译太难了

题意：给了n个数据，每个数据有一个值mi。又给了k个限制c。要把n个数据分成最小的组，满足每组中，大于i的数量不能超过ci个，问应该怎么分。

思路:先把n个数据都放到一组里，利用后缀和求大于i的数量有多少个，然后求出最少应该分len组。贪心从大到小模拟依次放入第0，1..len组中即可。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
ll a[maxn],c[maxn],cun[maxn],cnt[maxn];
    vector<int>ans[maxn];
int main()
{
    //freopen("input.txt", "r", stdin);
    int n,k;
    cin>>n>>k;
    rep(i,1,n){
        cin>>a[i];
        ++cnt[a[i]];
    }
    rep(i,1,k){
        cin>>c[i];
    }
    ll sum = 0;
    ll maxx = 1;

    repd(i,k,1){
        sum += cnt[i];
        if(sum % c[i] == 0){
            maxx = max(maxx,sum/c[i]);
        }
        else maxx = max(maxx,sum/c[i]+1);
    }
    int tot = 0;
    repd(i,k,1){
        while(cnt[i]){
            ans[tot%maxx].pb(i);
            cnt[i]--;
            tot++;
        }
    }
    cout<<maxx<<endl;
    for(int i = 0;i<maxx;++i){
        cout<<ans[i].size()<<sp;
        for(int j = 0 ;j < ans[i].size();++j){
            cout<<ans[i][j]<<sp;
        }
        cout<<endl;
    }    
}