B1. K for the Price of One (Easy Version)-----思维/排序/前缀和

This is the easy version of this problem. The only difference is the constraint on k — the number of gifts in the offer. In this version: k=2.

Vasya came to the store to buy goods for his friends for the New Year. It turned out that he was very lucky — today the offer “k of goods for the price of one” is held in store. Remember, that in this problem k=2.

Using this offer, Vasya can buy exactly k of any goods, paying only for the most expensive of them. Vasya decided to take this opportunity and buy as many goods as possible for his friends with the money he has.

More formally, for each good, its price is determined by ai — the number of coins it costs. Initially, Vasya has p coins. He wants to buy the maximum number of goods. Vasya can perform one of the following operations as many times as necessary:

Vasya can buy one good with the index i if he currently has enough coins (i.e p≥ai). After buying this good, the number of Vasya’s coins will decrease by ai, (i.e it becomes p:=p−ai).
Vasya can buy a good with the index i, and also choose exactly k−1 goods, the price of which does not exceed ai, if he currently has enough coins (i.e p≥ai). Thus, he buys all these k goods, and his number of coins decreases by ai (i.e it becomes p:=p−ai).
Please note that each good can be bought no more than once.

For example, if the store now has n=5 goods worth a1=2,a2=4,a3=3,a4=5,a5=7, respectively, k=2, and Vasya has 6 coins, then he can buy 3 goods. A good with the index 1 will be bought by Vasya without using the offer and he will pay 2 coins. Goods with the indices 2 and 3 Vasya will buy using the offer and he will pay 4 coins. It can be proved that Vasya can not buy more goods with six coins.

Help Vasya to find out the maximum number of goods he can buy.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases in the test.

The next lines contain a description of t test cases.

The first line of each test case contains three integers n,p,k (2≤n≤2⋅105, 1≤p≤2⋅109, k=2) — the number of goods in the store, the number of coins Vasya has and the number of goods that can be bought by the price of the most expensive of them.

The second line of each test case contains n integers ai (1≤ai≤104) — the prices of goods.

It is guaranteed that the sum of n for all test cases does not exceed 2⋅105. It is guaranteed that in this version of the problem k=2 for all test cases.

Output
For each test case in a separate line print one integer m — the maximum number of goods that Vasya can buy.

Example
inputCopy
6
5 6 2
2 4 3 5 7
5 11 2
2 4 3 5 7
2 10000 2
10000 10000
2 9999 2
10000 10000
5 13 2
8 2 8 2 5
3 18 2
1 2 3
outputCopy
3
4
2
0
4
3

题意:商店买东西，商店有n个物品，每个物品有自己的价格，商店有个优惠活动.
当你买恰好k个东西时可以只为其中最贵的那个付款.
求有限的钱中买到的最多的物品数量，你可以多次使用优惠。

解析:先排个序，求一个前缀和。当你购买第i件物品时，发现你只需要支付sum[i-k]+a[i]
为什么是i-k.因为可以赠送k-1个。
所以直接贪心只需要支付当前的价钱，和k-1个之后的价钱已经用前缀和维护好了
sum[i]=sum[i-k]+a[i]

通用B2

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+100;
int a[N],n,m,k;
int res[N];
int f[N];
int t;
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>n>>m>>k;
		memset(f,0,sizeof f);
		memset(res,0,sizeof res);
		for(int i=1;i<=n;i++) cin>>a[i];
		sort(a+1,a+1+n);
		for(int i=1;i<=n;i++) res[i]=res[i-1]+a[i];
		for(int i=k;i<=n;i++) res[i]=res[i-k]+a[i];
		int p;
		for(int i=0;i<=n;i++) if(m>=res[i])  p=i;
		cout<<p<<endl;
	}
}