ac

4 3 2
5 4 1 2
3 5 3 2

11

初始资金为3，最多做两个项目，每个项目的启动资金与利润见costs和profits。最优选择为：先做2号项目，做完之后资金增长到6,。然后做1号项目，做完之后资金增长到11。其他的任何选择都不会比这种选择好，所以返回11

import java.util.PriorityQueue;
import java.util.Scanner;



class Pro {
    public int cost;
    public int profit;
}

public class Main {

    public static Scanner sc = new Scanner(System.in);

    static int mod = 1_000_000_000 + 7;

    /**
     *
     * @param args
     */
    public static void main(String[] args) {
        int n = sc.nextInt();
        long w = sc.nextLong();
        int k = sc.nextInt();
        Pro[] pros = new Pro[n];
        for(int i=0;i<n;i++){
            pros[i] = new Pro();
            int cost = sc.nextInt();
            pros[i].cost = cost;
        }
        for(int i=0;i<n;i++){
            int profit = sc.nextInt();
            pros[i].profit = profit;
        }

        // profit 按从大到小排列
        PriorityQueue<Pro> priorityQueueProfit = new PriorityQueue<>((a,b)->{
            if(a.profit > b.profit){
                return -1;
            }else if(a.profit < b.profit){
                return 1;
            }
            return 0;
        });
        // cost 小的按顺序拍起来
        PriorityQueue<Pro> priorityQueueCost = new PriorityQueue<>((a,b)->{
            if(a.cost < b.cost){
                return -1;
            }else if(a.cost > b.cost){
                return 1;
            }
            return 0;
        });
        for(int i=0;i<n;i++){
            if(pros[i].cost <= w){
                priorityQueueProfit.add(pros[i]);
            }else{
                priorityQueueCost.add(pros[i]);
            }
        }
        while (k > 0){
            while (! priorityQueueCost.isEmpty() && priorityQueueCost.peek().cost <= w){
                Pro pro = priorityQueueCost.poll();
                priorityQueueProfit.add(pro);
            }
            if(priorityQueueProfit.isEmpty()){
               break;
            }
            Pro pro = priorityQueueProfit.poll();
            w += pro.profit;
            k--;
        }
        System.out.println(w);
    }


}
/*
4 3 2
5 4 1 2
3 5 3 2

11
*/

import java.util.PriorityQueue;
import java.util.Scanner;



class Pro {
    public long cost;
    public long profit;
}

public class Main {

    public static Scanner sc = new Scanner(System.in);

    static int mod = 1_000_000_000 + 7;

    static int N = 100_000;
    static int[] cost = new int[N];
    static int[] profit = new int[N];
    /**
     *
     * @param args
     */
    public static void main(String[] args) {
        int n = sc.nextInt();
        long w = sc.nextLong();
        int k = sc.nextInt();

        for(int i=0;i<n;i++){
            cost[i] = sc.nextInt();
        }
        for(int i=0;i<n;i++){
            profit[i] = sc.nextInt();
        }

        // profit 按从大到小排列
        PriorityQueue<Integer> priorityQueueProfit = new PriorityQueue<>((a,b)->{
            if(profit[a] > profit[b]){
                return -1;
            }else if(profit[a] < profit[b]){
                return 1;
            }
            return 0;
        });
        // cost 小的按顺序拍起来
        PriorityQueue<Integer> priorityQueueCost = new PriorityQueue<>((a,b)->{
            if(cost[a] < cost[b]){
                return -1;
            }else if(cost[a] > cost[b]){
                return 1;
            }
            return 0;
        });
        for(int i=0;i<n;i++){
            if(cost[i] <= w){
                priorityQueueProfit.add(i);
            }else{
                priorityQueueCost.add(i);
            }
        }
        while (k != 0){
            while (! priorityQueueCost.isEmpty() && cost[priorityQueueCost.peek()] <= w){
                int index = priorityQueueCost.poll();
                priorityQueueProfit.add(index);
            }
            if(priorityQueueProfit.isEmpty()){
                break;
            }
            w +=  profit[priorityQueueProfit.poll()];
            k--;
        }
        System.out.println(w);
    }


}
/*
4 3 2
5 4 1 2
3 5 3 2

11
*/